splay

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这个星期打了一些splay，下面在这里做下总结吧。

splay 的核心要实现的就是把二叉排序树中的一个节点移到它的某个祖先节点的位置。

这个操作初看来可能没有什么实际的用途，但是仔细思考简直是非常方便。

当我们需要对一个区间进行操作时，只要把这个区间的左端点的前一个点和右端点的后一个点分别旋到根节点和根节点的右儿子节点，这时候一个感人肺腑的现象就出现了：以根节点的右儿子节点的左儿子节点为根的这棵树就是所询问的线段。画个图是很显然的。

有了这个操作后很多东西就都可以迎刃而解了。

其实splay比线段树好的地方大概也就是这里，即它对区间的灵活性。

总结splay可以多实现的东西我目前大概只知道这三个：插入一个区间，删除一个区间，翻转一个区间。

下面看实现，

最暴力的方法就是把以目标节点为根的子树拆了然后从新建，但是容易发现有一种更加便捷的方法就是把当前节点一层一层地旋上去。

这里使用的是双旋，所以分为两种情况：父亲就是目标节点，直接转上去就好了；父亲不是目标节点，根据当前节点和当前节点的父节点是否同是自身父节点的左（右）儿子再分成两种情况讨论：如果左右性相同，先把父节点向上转，再把当前节点向上转；如果左右性不同，就把当前节点向上转两次。（转法不唯一，只要可以把当前节点转到父节点的父节点就可以了）。

下面处理刚才说的那个旋转。如果把一个点向右上旋，当前节点右儿子的位置肯定就会被当前节点的父节点占领，哪怎么办呢？显然这里多出来了一个点就一定有一个地方少了一个点—— 原来的父节点的左儿子（即当前节点）现在没有了，于是自然要把多出来的这个接上去，考虑二叉排序树的性质这样接时没有问题的。显然在旋转后，当前节点变成了原先的父节点的父节点，变成了原先的父节点的父节点的子节点，所以再把这两层关系分别处理一下就好了。

以上就是splay全部的核心思想了，具体实现时考虑左右的对称性可以把左旋和右旋放到一个函数里写，一个 k 等于1时是右旋，等于0时是左旋。

实现起来极其简单。只有简洁优美的十几行。

void jie(int x, int y, int k){if(y)ch[y][k] = x; fat[x] = y;}void rotate(int x, int k){    int y = fat[x];    jie(ch[x][k], y, !k),jie(x, fat[y], ch[fat[y]][1] == y),jie(y, x, k);}void splay(int x, int goal){    while(fat[x] != goal){        if(fat[fat[x]] == goal){rotate(x, ch[fat[x]][0] == x); break;}        int y = fat[x], ky = ch[fat[y]][1] == y;        if(ch[y][ky] == x)rotate(y, !ky), rotate(x, !ky);        else rotate(x, ky), rotate(x, !ky);    }    if(!goal)root = x;}

虽然splay操作真的很短，但是考虑到简单的操作用线段树就可以了，所以splay的题一般会出很多种操作，分一下类就会觉得代码很长，其实没有什么的，也就是维护一个二叉排序树的各种性质。

先上几道可以用线段树等较简单方法实现的题来巩固对splay的理解。

hdu1754 I hate it

这道题相信没人没有用线段树做过，因为什么操作都没有所以用splay打也超级短啊! 短的令人想哭。

<pre style="font-family:Courier New;text-align:left;"><span style="color:blue;">#include <iostream>#include <cstdio>#define MAXN 200005</span><strong><span style="color:#0000FF;">using namespace</span></strong> std<strong><span style="color:#FF00FF;">;</span><span style="color:blue;">int</span></strong> n<strong><span style="color:#FF00FF;">,</span></strong> q<strong><span style="color:#FF00FF;">,</span></strong> num<strong><span style="color:#FF00FF;">[</span></strong>MAXN<strong><span style="color:#FF00FF;">],</span></strong> maxx<strong><span style="color:#FF00FF;">[</span></strong>MAXN<strong><span style="color:#FF00FF;">],</span></strong> fat<strong><span style="color:#FF00FF;">[</span></strong>MAXN<strong><span style="color:#FF00FF;">],</span></strong> root<strong><span style="color:#FF00FF;">,</span></strong> ch<strong><span style="color:#FF00FF;">[</span></strong>MAXN<strong><span style="color:#FF00FF;">][</span></strong><span style="color:#CC3300;">2</span><strong><span style="color:#FF00FF;">];</span><span style="color:blue;">void</span></strong> pushup<strong><span style="color:#FF00FF;">(</span><span style="color:blue;">int</span></strong> x<strong><span style="color:#FF00FF;">){</span></strong>maxx<strong><span style="color:#FF00FF;">[</span></strong>x<strong><span style="color:#FF00FF;">] =</span></strong> max<strong><span style="color:#FF00FF;">(</span></strong>num<strong><span style="color:#FF00FF;">[</span></strong>x<strong><span style="color:#FF00FF;">],</span></strong> max<strong><span style="color:#FF00FF;">(</span></strong>maxx<strong><span style="color:#FF00FF;">[</span></strong>ch<strong><span style="color:#FF00FF;">[</span></strong>x<strong><span style="color:#FF00FF;">][</span></strong><span style="color:#CC3300;">1</span><strong><span style="color:#FF00FF;">]],</span></strong> maxx<strong><span style="color:#FF00FF;">[</span></strong>ch<strong><span style="color:#FF00FF;">[</span></strong>x<strong><span style="color:#FF00FF;">][</span></strong><span style="color:#CC3300;">0</span><strong><span style="color:#FF00FF;">]]));}</span><span style="color:blue;">void</span></strong> jie<strong><span style="color:#FF00FF;">(</span><span style="color:blue;">int</span></strong> x<strong><span style="color:#FF00FF;">,</span><span style="color:blue;"> int</span></strong> y<strong><span style="color:#FF00FF;">,</span><span style="color:blue;"> int</span></strong> k<strong><span style="color:#FF00FF;">){</span><span style="color:#0000FF;">if</span><span style="color:#FF00FF;">(</span></strong>y<strong><span style="color:#FF00FF;">)</span></strong>ch<strong><span style="color:#FF00FF;">[</span></strong>y<strong><span style="color:#FF00FF;">][</span></strong>k<strong><span style="color:#FF00FF;">] =</span></strong> x<strong><span style="color:#FF00FF;">;</span></strong> fat<strong><span style="color:#FF00FF;">[</span></strong>x<strong><span style="color:#FF00FF;">] =</span></strong> y<strong><span style="color:#FF00FF;">;}</span><span style="color:blue;">void</span></strong> rotate<strong><span style="color:#FF00FF;">(</span><span style="color:blue;">int</span></strong> x<strong><span style="color:#FF00FF;">,</span><span style="color:blue;"> int</span></strong> k<strong><span style="color:#FF00FF;">){</span><span style="color:blue;">    int</span></strong> y<strong><span style="color:#FF00FF;"> =</span></strong> fat<strong><span style="color:#FF00FF;">[</span></strong>x<strong><span style="color:#FF00FF;">];</span></strong>    jie<strong><span style="color:#FF00FF;">(</span></strong>ch<strong><span style="color:#FF00FF;">[</span></strong>x<strong><span style="color:#FF00FF;">][</span></strong>k<strong><span style="color:#FF00FF;">],</span></strong> y<strong><span style="color:#FF00FF;">, !</span></strong>k<strong><span style="color:#FF00FF;">),</span></strong>jie<strong><span style="color:#FF00FF;">(</span></strong>x<strong><span style="color:#FF00FF;">,</span></strong> fat<strong><span style="color:#FF00FF;">[</span></strong>y<strong><span style="color:#FF00FF;">],</span></strong> ch<strong><span style="color:#FF00FF;">[</span></strong>fat<strong><span style="color:#FF00FF;">[</span></strong>y<strong><span style="color:#FF00FF;">]][</span></strong><span style="color:#CC3300;">1</span><strong><span style="color:#FF00FF;">] ==</span></strong> y<strong><span style="color:#FF00FF;">),</span></strong>jie<strong><span style="color:#FF00FF;">(</span></strong>y<strong><span style="color:#FF00FF;">,</span></strong> x<strong><span style="color:#FF00FF;">,</span></strong> k<strong><span style="color:#FF00FF;">);</span></strong>    pushup<strong><span style="color:#FF00FF;">(</span></strong>y<strong><span style="color:#FF00FF;">);}</span><span style="color:blue;">void</span></strong> splay<strong><span style="color:#FF00FF;">(</span><span style="color:blue;">int</span></strong> x<strong><span style="color:#FF00FF;">,</span><span style="color:blue;"> int</span></strong> goal<strong><span style="color:#FF00FF;">){</span><span style="color:#0000FF;">    while</span><span style="color:#FF00FF;">(</span></strong>fat<strong><span style="color:#FF00FF;">[</span></strong>x<strong><span style="color:#FF00FF;">] !=</span></strong> goal<strong><span style="color:#FF00FF;">){</span><span style="color:#0000FF;">        if</span><span style="color:#FF00FF;">(</span></strong>fat<strong><span style="color:#FF00FF;">[</span></strong>fat<strong><span style="color:#FF00FF;">[</span></strong>x<strong><span style="color:#FF00FF;">]] ==</span></strong> goal<strong><span style="color:#FF00FF;">){</span></strong>rotate<strong><span style="color:#FF00FF;">(</span></strong>x<strong><span style="color:#FF00FF;">,</span></strong> ch<strong><span style="color:#FF00FF;">[</span></strong>fat<strong><span style="color:#FF00FF;">[</span></strong>x<strong><span style="color:#FF00FF;">]][</span></strong><span style="color:#CC3300;">0</span><strong><span style="color:#FF00FF;">] ==</span></strong> x<strong><span style="color:#FF00FF;">);</span><span style="color:#0000FF;"> break</span><span style="color:#FF00FF;">;}</span><span style="color:blue;">        int</span></strong> y<strong><span style="color:#FF00FF;"> =</span></strong> fat<strong><span style="color:#FF00FF;">[</span></strong>x<strong><span style="color:#FF00FF;">],</span></strong> ky<strong><span style="color:#FF00FF;"> =</span></strong> ch<strong><span style="color:#FF00FF;">[</span></strong>fat<strong><span style="color:#FF00FF;">[</span></strong>y<strong><span style="color:#FF00FF;">]][</span></strong><span style="color:#CC3300;">1</span><strong><span style="color:#FF00FF;">] ==</span></strong> y<strong><span style="color:#FF00FF;">;</span><span style="color:#0000FF;">        if</span><span style="color:#FF00FF;">(</span></strong>ch<strong><span style="color:#FF00FF;">[</span></strong>y<strong><span style="color:#FF00FF;">][</span></strong>ky<strong><span style="color:#FF00FF;">] ==</span></strong> x<strong><span style="color:#FF00FF;">)</span></strong>rotate<strong><span style="color:#FF00FF;">(</span></strong>y<strong><span style="color:#FF00FF;">, !</span></strong>ky<strong><span style="color:#FF00FF;">),</span></strong> rotate<strong><span style="color:#FF00FF;">(</span></strong>x<strong><span style="color:#FF00FF;">, !</span></strong>ky<strong><span style="color:#FF00FF;">);</span><span style="color:#0000FF;">        else</span></strong> rotate<strong><span style="color:#FF00FF;">(</span></strong>x<strong><span style="color:#FF00FF;">,</span></strong> ky<strong><span style="color:#FF00FF;">),</span></strong> rotate<strong><span style="color:#FF00FF;">(</span></strong>x<strong><span style="color:#FF00FF;">, !</span></strong>ky<strong><span style="color:#FF00FF;">);    }</span></strong>pushup<strong><span style="color:#FF00FF;">(</span></strong>x<strong><span style="color:#FF00FF;">);</span><span style="color:#0000FF;">    if</span><span style="color:#FF00FF;">(!</span></strong>goal<strong><span style="color:#FF00FF;">)</span></strong>root<strong><span style="color:#FF00FF;"> =</span></strong> x<strong><span style="color:#FF00FF;">;}</span><span style="color:blue;">int</span><span style="color:#0000FF;"> main</span><span style="color:#FF00FF;">(){</span><span style="color:#0000FF;">    while</span><span style="color:#FF00FF;">(</span></strong>scanf<strong><span style="color:#FF00FF;">(</span></strong><span style="color:green;">"%d%d"</span><strong><span style="color:#FF00FF;">, &</span></strong>n<strong><span style="color:#FF00FF;">, &</span></strong>q<strong><span style="color:#FF00FF;">) !=</span></strong> EOF<strong><span style="color:#FF00FF;">){</span><span style="color:#0000FF;">        for</span><span style="color:#FF00FF;">(</span><span style="color:blue;">int</span></strong> i<strong><span style="color:#FF00FF;"> =</span></strong><span style="color:#CC3300;"> 2</span><strong><span style="color:#FF00FF;">;</span></strong> i<strong><span style="color:#FF00FF;"> <=</span></strong> n<strong><span style="color:#FF00FF;"> +</span></strong><span style="color:#CC3300;"> 1</span><strong><span style="color:#FF00FF;">;</span></strong> i<strong><span style="color:#FF00FF;"> ++)</span></strong>scanf<strong><span style="color:#FF00FF;">(</span></strong><span style="color:green;">"%d"</span><strong><span style="color:#FF00FF;">, &</span></strong>num<strong><span style="color:#FF00FF;">[</span></strong>i<strong><span style="color:#FF00FF;">]);</span></strong> num<strong><span style="color:#FF00FF;">[</span></strong><span style="color:#CC3300;">1</span><strong><span style="color:#FF00FF;">] =</span></strong> num<strong><span style="color:#FF00FF;">[</span></strong>n<strong><span style="color:#FF00FF;"> +</span></strong><span style="color:#CC3300;"> 2</span><strong><span style="color:#FF00FF;">] = -</span></strong><span style="color:#CC3300;">100000000</span><strong><span style="color:#FF00FF;">;</span></strong> maxx<strong><span style="color:#FF00FF;">[</span></strong>n<strong><span style="color:#FF00FF;"> +</span></strong><span style="color:#CC3300;"> 3</span><strong><span style="color:#FF00FF;">] =</span></strong><span style="color:#CC3300;"> 0</span><strong><span style="color:#FF00FF;">;</span><span style="color:#0000FF;">        for</span><span style="color:#FF00FF;">(</span><span style="color:blue;">int</span></strong> i<strong><span style="color:#FF00FF;"> =</span></strong> n<strong><span style="color:#FF00FF;"> +</span></strong><span style="color:#CC3300;"> 2</span><strong><span style="color:#FF00FF;">;</span></strong> i<strong><span style="color:#FF00FF;">;</span></strong> i<strong><span style="color:#FF00FF;"> --){</span></strong>            ch<strong><span style="color:#FF00FF;">[</span></strong>i<strong><span style="color:#FF00FF;">][</span></strong><span style="color:#CC3300;">0</span><strong><span style="color:#FF00FF;">] =</span></strong><span style="color:#CC3300;"> 0</span><strong><span style="color:#FF00FF;">; (</span></strong>i<strong><span style="color:#FF00FF;"> ==</span></strong> n<strong><span style="color:#FF00FF;"> +</span></strong><span style="color:#CC3300;"> 2</span><strong><span style="color:#FF00FF;"> )?</span></strong> ch<strong><span style="color:#FF00FF;">[</span></strong>i<strong><span style="color:#FF00FF;">][</span></strong><span style="color:#CC3300;">1</span><strong><span style="color:#FF00FF;">] =</span></strong><span style="color:#CC3300;"> 0</span><strong><span style="color:#FF00FF;"> :</span></strong> ch<strong><span style="color:#FF00FF;">[</span></strong>i<strong><span style="color:#FF00FF;">][</span></strong><span style="color:#CC3300;">1</span><strong><span style="color:#FF00FF;">] =</span></strong> i<strong><span style="color:#FF00FF;"> +</span></strong><span style="color:#CC3300;"> 1</span><strong><span style="color:#FF00FF;">;</span></strong>            fat<strong><span style="color:#FF00FF;">[</span></strong>i<strong><span style="color:#FF00FF;">] =</span></strong> i<strong><span style="color:#FF00FF;"> -</span></strong><span style="color:#CC3300;"> 1</span><strong><span style="color:#FF00FF;">;</span></strong> maxx<strong><span style="color:#FF00FF;">[</span></strong>i<strong><span style="color:#FF00FF;">] =</span></strong> max<strong><span style="color:#FF00FF;">(</span></strong>maxx<strong><span style="color:#FF00FF;">[</span></strong>i<strong><span style="color:#FF00FF;"> +</span></strong><span style="color:#CC3300;"> 1</span><strong><span style="color:#FF00FF;">],</span></strong> num<strong><span style="color:#FF00FF;">[</span></strong>i<strong><span style="color:#FF00FF;">]);        }</span></strong> root<strong><span style="color:#FF00FF;"> =</span></strong><span style="color:#CC3300;"> 1</span><strong><span style="color:#FF00FF;">;</span><span style="color:#0000FF;">         while</span><span style="color:#FF00FF;">(</span></strong>q<strong><span style="color:#FF00FF;"> --){</span><span style="color:blue;">            char</span></strong> s<strong><span style="color:#FF00FF;">[</span></strong><span style="color:#CC3300;">5</span><strong><span style="color:#FF00FF;">];</span><span style="color:blue;"> int</span></strong> x<strong><span style="color:#FF00FF;">,</span></strong> y<strong><span style="color:#FF00FF;">;</span></strong> scanf<strong><span style="color:#FF00FF;">(</span></strong><span style="color:green;">"%s%d%d"</span><strong><span style="color:#FF00FF;">,</span></strong> s<strong><span style="color:#FF00FF;">, &</span></strong>x<strong><span style="color:#FF00FF;">, &</span></strong>y<strong><span style="color:#FF00FF;">);</span></strong>x<strong><span style="color:#FF00FF;"> ++;</span><span style="color:#0000FF;">            if</span><span style="color:#FF00FF;">(</span></strong>s<strong><span style="color:#FF00FF;">[</span></strong><span style="color:#CC3300;">0</span><strong><span style="color:#FF00FF;">] ==</span></strong><span style="color:green;"> 'U'</span><strong><span style="color:#FF00FF;">)</span></strong>splay<strong><span style="color:#FF00FF;">(</span></strong>x<strong><span style="color:#FF00FF;">,</span></strong><span style="color:#CC3300;"> 0</span><strong><span style="color:#FF00FF;">),</span></strong>num<strong><span style="color:#FF00FF;">[</span></strong>x<strong><span style="color:#FF00FF;">] =</span></strong> y<strong><span style="color:#FF00FF;">,</span></strong> pushup<strong><span style="color:#FF00FF;">(</span></strong>x<strong><span style="color:#FF00FF;">);</span><span style="color:#0000FF;">            else</span></strong> splay<strong><span style="color:#FF00FF;">(</span></strong>x<strong><span style="color:#FF00FF;"> -</span></strong><span style="color:#CC3300;"> 1</span><strong><span style="color:#FF00FF;">,</span></strong><span style="color:#CC3300;"> 0</span><strong><span style="color:#FF00FF;">),</span></strong> splay<strong><span style="color:#FF00FF;">(</span></strong>y<strong><span style="color:#FF00FF;"> +</span></strong><span style="color:#CC3300;"> 2</span><strong><span style="color:#FF00FF;">,</span></strong> x<strong><span style="color:#FF00FF;"> -</span></strong><span style="color:#CC3300;"> 1</span><strong><span style="color:#FF00FF;">),</span></strong> printf<strong><span style="color:#FF00FF;">(</span></strong><span style="color:green;">"%d\n"</span><strong><span style="color:#FF00FF;">,</span></strong> maxx<strong><span style="color:#FF00FF;">[</span></strong>ch<strong><span style="color:#FF00FF;">[</span></strong>y<strong><span style="color:#FF00FF;"> +</span></strong><span style="color:#CC3300;"> 2</span><strong><span style="color:#FF00FF;">][</span></strong><span style="color:#CC3300;">0</span><strong><span style="color:#FF00FF;">]]);        }            }</span><span style="color:#0000FF;">    return</span></strong><span style="color:#CC3300;"> 0</span><strong><span style="color:#FF00FF;">;}</span></strong>

poj 3468 a simple promble with integers

同裸线段树经典题。多了一个修改操作。

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#define MAXN 100005using namespace std;int n, q, root, num[MAXN], ch[MAXN][2], fat[MAXN], r[MAXN], sz[MAXN];long long summ[MAXN];inline void jie(int x, int y, int k){if(y)ch[y][k] = x; fat[x] = y;}inline void pushup(int x){sz[x] = sz[ch[x][1]] + sz[ch[x][0]] + 1;summ[x] = r[x] + num[x] + summ[ch[x][0]] + summ[ch[x][1]];}inline void pushdown(int x){if(!r[x])return;num[x] += r[x];r[ch[x][0]] += r[x]; r[ch[x][1]] += r[x];summ[ch[x][0]] += (long long)sz[ch[x][0]] * r[x];summ[ch[x][1]] += (long long)sz[ch[x][1]] * r[x];r[x] = 0;}inline void rotate(int x, int k){int y = fat[x]; pushdown(y); pushdown(x);jie(ch[x][k], y, !k), jie(x, fat[y], ch[fat[y]][1] == y), jie(y, x, k);pushup(y); }inline void splay(int x, int goal){pushdown(x);while(fat[x] != goal){if(fat[fat[x]] == goal){rotate(x, ch[fat[x]][0] == x); break;}int y = fat[x], ky = ch[fat[y]][1] == y;if(ch[y][ky] == x)rotate(y, !ky), rotate(x, !ky);else rotate(x, ky), rotate(x, !ky);} pushup(x);if(! goal)root = x;}int main(){scanf("%d%d", &n, &q);for(int i = 2; i <= n + 1; i ++)scanf("%d", &num[i]); fat[n + 2] = n + 1;for(int i = n + 1; i; i --){ch[i][1] = i + 1; summ[i] = summ[i + 1] + num[i]; sz[i] = sz[i + 1] + 1; fat[i] = i - 1;}root = 1;while(q --){char s[5]; int x, y, z; scanf("%s%d%d", s, &x, &y);x ++; y ++;if(s[0] == 'Q')splay(x - 1, 0), splay(y + 1, x - 1), cout<<summ[ch[y + 1][0]]<<endl;else{scanf("%d", &z);splay(x - 1, 0); splay(y + 1, x - 1);r[ch[y + 1][0]] += z;summ[ch[y + 1][0]] += (long long)z * sz[ch[y + 1][0]];}}return 0;}

bzoj 1588 营业额统计

所以这次又新增了一个插入操作。

#include <iostream>#include <cstdio>#include <cmath>#include <algorithm>#include <cstring>#define MAXN 100005#define INF 1<<30using namespace std;int n, num, ch[MAXN][2], a[MAXN], cnt, root, ans, fat[MAXN];void addnode(int father, int k){    a[++ cnt] = k;    fat[cnt] = father;}void rotate(int x, int kind){    int f = fat[x];    ch[f][!kind] = ch[x][kind];    fat[ch[x][kind]] = f;    //if(fat[f])    ch[fat[f]][ch[fat[f]][1] == f] = x;    fat[x] = fat[f];    ch[x][kind] = f;    fat[f] = x;}void splay(int r){    while(fat[r]){        if(!fat[fat[r]]){rotate(r, ch[fat[r]][1] != r); continue;}        int f = fat[r];        int kindf = ch[fat[f]][1] == f;        if(ch[f][kindf] == r)rotate(f, !kindf), rotate(r, !kindf);        else rotate(r, kindf), rotate(r, !kindf);    } root = r;}int insert(int k){    int r = root;    while(ch[r][a[r] < k]){        if(a[r] == k){splay(r); return 0;}        r = ch[r][a[r] < k];    }    ch[r][a[r] < k] = cnt + 1;    addnode(r, k);    splay(cnt); return 1;}int get(int x, int k){    int tmp = ch[x][!k];    if(!tmp)return INF;    while(ch[tmp][k])tmp = ch[tmp][k];    return abs(a[tmp] - a[x]);}int main(){    //freopen("1588.in", "r", stdin);    scanf("%d%d", &n, &num);    root = 1; addnode(0, num); ans += num;    for(int i = 2; i <= n; i ++){        num = 0; scanf("%d", &num);        if(! insert(num))continue;        ans += min(get(root, 0), get(root, 1));    }cout<<ans<<endl;    return 0;}

bzoj1503 郁闷的出纳员

所以这次又新增了删除操作。

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#define MAXN 200005using namespace std;int n, minn, minn0, x, sz[MAXN], ch[MAXN][2], pre[MAXN], cnt[MAXN], val[MAXN], summ, rt, top;inline void pushup(int x){sz[x] = cnt[x] + sz[ch[x][0]] + sz[ch[x][1]];}inline void jie(int y, int x, int k){if(y)ch[y][k] = x; pre[x] = y;}inline void rotate(int x, int k){   int y = pre[x];jie(y, ch[x][k], !k);    jie(pre[y], x, ch[pre[y]][1] == y);    jie(x, y, k);    pushup(y);}inline void splay(int x, int goal){    while(pre[x] != goal){        if(pre[pre[x]] == goal){rotate(x, ch[pre[x]][0] == x); break;}        int y = pre[x], kf = ch[pre[y]][1] == y;        if(ch[y][kf] == x)rotate(y, !kf), rotate(x, !kf);        else rotate(x, kf), rotate(x, !kf);    } pushup(x);    if(!goal)rt = x;}inline void newnode(int &x, int c){    x = ++ top;    sz[x] = 1; cnt[x] = 1; val[x] = c;    }inline void insert(int &x, int key, int f){    if(!x){        newnode(x, key);        pre[x] = f;        splay(x, 0); return;        }        if(key == val[x]){        cnt[x] ++; sz[x] ++;        splay(x, 0); return ;        }    if(key < val[x])insert(ch[x][0], key, x);    else insert(ch[x][1], key, x);    pushup(x);}inline void del(int &x, int f){    if(!x)return;    if(val[x] >= minn) del(ch[x][0], x);    else{        summ += sz[ch[x][0]] + cnt[x];        x = ch[x][1];        pre[x] = f;        if(f == 0)rt = x;        del(x, f);        }    if(x)pushup(x);}inline int findk(int x, int k){//cout<<x<<endl;    if(k < sz[ch[x][0]] + 1)return findk(ch[x][0], k);    if(k > sz[ch[x][0]] + cnt[x])return findk(ch[x][1], k - sz[ch[x][0]] - cnt[x]);    splay(x, 0); return val[x];    }int main(){    scanf("%d%d", &n, &minn); minn0 = minn;    char s[10]; int k;     while(n --){        scanf("%s%d", s, &k);        if(s[0] == 'A')minn -= k;        if(s[0] == 'S'){            minn += k;            if(sz[rt])del(rt, 0);            }            if(s[0] == 'I' && k >= minn0)insert(rt, k + minn - minn0, 0);        if(s[0] == 'F'){            if(k > sz[rt])puts("-1");            else printf("%d\n", findk(rt, sz[rt] - k + 1) - minn + minn0);            }    }cout<<summ<<endl;    return 0;    }

bzoj1798 维护序列

这时候有了修改操作，自然要用到懒标记，懒标记的pushup 和pushdown和线段树基本相同。

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#define MAXN 100005using namespace std;int n, p, la, a, b, c;struct Tree{    int l, r;long long plus, mul, summ;}tree[MAXN * 4];inline void pushup(int t){tree[t].summ = (tree[t + t].summ + tree[t + t + 1].summ) % p;}inline void pushdown(int t){    tree[t + t].plus = (tree[t + t].plus * tree[t].mul + tree[t].plus) % p;    tree[t + t + 1].plus = (tree[t + t + 1].plus * tree[t].mul + tree[t].plus) % p;    tree[t + t].mul = tree[t + t].mul * tree[t].mul % p;    tree[t + t + 1].mul = tree[t + t + 1].mul * tree[t].mul % p;    tree[t + t].summ = (tree[t + t].summ * tree[t].mul + tree[t].plus * (tree[t + t].r - tree[t + t].l + 1)) % p;    tree[t + t + 1].summ = (tree[t + t + 1].summ * tree[t].mul + tree[t].plus * (tree[t + t + 1].r - tree[t + t + 1].l + 1)) % p;    tree[t].plus = 0; tree[t].mul = 1;}void build(int l, int r, int t){    tree[t].l = l; tree[t].r = r;    tree[t].plus = 0; tree[t].mul = 1;    if(l == r){scanf("%lld", &tree[t].summ); return;}    int mid = l + r >> 1;    build(l, mid, t + t); build(mid + 1, r, t + t + 1);    pushup(t);}void alter(int t, int l, int r){    if(tree[t].l >= l && tree[t].r <= r){        if(la == 1){            tree[t].plus = tree[t].plus * c % p;            tree[t].mul = tree[t].mul * c % p;            tree[t].summ = tree[t].summ * c % p;        }else{            tree[t].plus = (tree[t].plus + c) % p;            tree[t].summ = (tree[t].summ + (long long)c * (tree[t].r - tree[t].l + 1)) % p;        }        return;    }pushdown(t);    int mid = tree[t].l + tree[t].r >> 1;    if(mid >= l)alter(t + t, l, r);    if(mid  < r)alter(t + t + 1, l, r);    pushup(t);}long long query(int t, int l, int r){    if(tree[t].l >= l && tree[t].r <= r){        return tree[t].summ % p;    }pushdown(t);    int mid = tree[t].l + tree[t].r >> 1;long long cnt = 0;    if(mid >= l)cnt = query(t + t, l, r);    if(mid < r)cnt += query(t + t + 1, l, r);    pushup(t);    return cnt % p;}int main(){    scanf("%d%d", &n, &p);    build(1, n, 1);    int m; scanf("%d", &m); while(m --){        scanf("%d%d%d", &la, &a, &b);        if(la != 3){scanf("%d", &c);            alter(1, a, b);        }else{            printf("%lld\n", query(1, a, b));        }    }    return 0;}

bzoj1500 维修数列

这道题要处理的东西有点多，但实际上每一个小模块都是之前写过的，只是把它们整合在一起而已。

易错点： 1，在转移的时候要判左右儿子是否为空（我开始是想不去特判而给空的节点赋一个很小的值什么的，但都不好实现）

2，垃圾回收避免MLE。

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#include <queue>#define MAXN 510005#define INF 1<<30using namespace std;int n, m, root, cnt, fat[MAXN], ch[MAXN][2], sz[MAXN], summ[MAXN], lmax[MAXN], rmax[MAXN], maxx[MAXN], num[MAXN], a[MAXN];bool rev[MAXN], change[MAXN];queue<int>q;void newnode(int &x, int fatt, int val){    if(!q.empty())x = q.front(), q.pop();    else x = ++ cnt;    fat[x] = fatt;    ch[x][0] = ch[x][1] = rev[x] = change[x] = 0;    sz[x] = 1;    summ[x] = lmax[x] = rmax[x] = maxx[x] = num[x] = val;}inline void pushdown(int x){    if(!x)return;    if(rev[x]){        rev[x] = 0;        if(ch[x][0])rev[ch[x][0]] ^= 1;        if(ch[x][1])rev[ch[x][1]] ^= 1;        swap(ch[x][0], ch[x][1]);        swap(lmax[x], rmax[x]);             }    if(change[x]){        change[x] = 0;        if(ch[x][0]){            change[ch[x][0]] = 1;            num[ch[x][0]] = num[x];            summ[ch[x][0]] = num[x] * sz[ch[x][0]];            rmax[ch[x][0]] = lmax[ch[x][0]] = maxx[ch[x][0]] = max(num[x], summ[ch[x][0]]);        }        if(ch[x][1]){            change[ch[x][1]] = 1;            num[ch[x][1]] = num[x];            summ[ch[x][1]] = num[x] * sz[ch[x][1]];            rmax[ch[x][1]] = lmax[ch[x][1]] = maxx[ch[x][1]] = max(num[x], summ[ch[x][1]]);        }    }}inline void pushup(int x){    if(!x)return;    pushdown(ch[x][0]), pushdown(ch[x][1]);    summ[x] = (ch[x][0] ? summ[ch[x][0]] : 0) + (ch[x][1] ? summ[ch[x][1]] : 0) + num[x];    sz[x] = (ch[x][0] ? sz[ch[x][0]] : 0) + (ch[x][1] ? sz[ch[x][1]] : 0) + 1;    lmax[x] = max(ch[x][0] ? lmax[ch[x][0]] : num[x], (ch[x][0] ? summ[ch[x][0]] : 0) + num[x] + max(0, ch[x][1] ? lmax[ch[x][1]] : 0));    rmax[x] = max(ch[x][1] ? rmax[ch[x][1]] : num[x], (ch[x][1] ? summ[ch[x][1]] : 0) + num[x] + max(0, ch[x][0] ? rmax[ch[x][0]] : 0));    maxx[x] = max(max(ch[x][0] ? maxx[ch[x][0]] : num[x], ch[x][1] ? maxx[ch[x][1]] : num[x]), max(0, ch[x][0] ? rmax[ch[x][0]] : 0) + max(0, ch[x][1] ? lmax[ch[x][1]] : 0) + num[x]);}void build(int &x, int l, int r, int fat){    if(l <= r){        int mid = l + r >> 1;        //cout<<mid<<endl;        //cout<<a[mid]<<endl;        newnode(x, fat, a[mid]);        if(l == r)return;        if(mid > l)build(ch[x][0], l, mid - 1, x);        build(ch[x][1], mid + 1, r, x);        pushup(x);    }}inline void jie(int x, int y, int k){if(y)ch[y][k] = x; fat[x] = y;}inline void rotate(int x, int k){    int y = fat[x];pushdown(y);//pushdown(x);    jie(ch[x][k], y, !k);    jie(x, fat[y], ch[fat[y]][1] == y);    jie(y, x, k);    pushup(y);}void splay(int x, int goal){    if(x == goal)return;    pushdown(x);    while(fat[x] != goal){        if(fat[fat[x]] == goal){        //  pushdown(fat[x]);        //  pushdown(x);            rotate(x, ch[fat[x]][0] == x); break;        }    //  pushdown(fat[fat[x]]);    //  pushdown(fat[x]);    //  pushdown(x);        int y = fat[x], ky = ch[fat[y]][1] == y;        if(ch[y][ky] == x)rotate(y, !ky), rotate(x, !ky);        else rotate(x, ky), rotate(x, !ky);    }pushup(x);    if(!goal)root = x;}int select(int k){    int x = root; pushdown(x);    while(sz[ch[x][0]] + 1 != k){        if(sz[ch[x][0]] + 1 < k){            k -= sz[ch[x][0]] + 1;            x = ch[x][1];        }        else x = ch[x][0];        pushdown(x);    }return x;}void Reverse(int a, int b){    int A = select(a), B = select(b + 2);    splay(A, 0), splay(B, A);    rev[ch[B][0]] ^= 1;    pushup(B), pushup(A);}int Get_sum(int a, int b){    int A = select(a), B = select(b + 2);    splay(A, 0), splay(B, A);    return summ[ch[B][0]]; }void del(int x){    if(!x)return;    q.push(x);    del(ch[x][0]);    del(ch[x][1]);}void Delete(int a, int b){    int A = select(a), B = select(b + 2);    splay(A, 0), splay(B, A);    del(ch[B][0]);    ch[B][0] = 0;    pushup(B), pushup(A);}int Max_sum(){    int A = select(1), B = select(sz[root]);    splay(A, 0), splay(B, A);    return maxx[ch[B][0]];}void Make_same(int a, int b, int c){    int A = select(a), B = select(b + 2);    splay(A, 0), splay(B, A);    int x = ch[B][0];    change[x] = 1;    num[x] = c;    summ[x] = c * sz[x];    rmax[x] = lmax[x] = maxx[x] = max(c, summ[x]);    pushup(B), pushup(A);}void Insert(int a, int b){a ++;    int A = select(a), B = select(a + 1);    splay(A, 0), splay(B, A);    //  printf("!! %d %d\n", summ[ch[ch[root][1]][0]], summ[root]);    build(ch[ch[root][1]][0], 1, b, ch[root][1]);    pushup(ch[root][1]); pushup(root);//  printf("!! %d %d\n", summ[ch[ch[root][1]][0]], summ[root]);}int main(){//  freopen("sequence.in", "r", stdin);//  freopen("sequence.out", "w", stdout);    scanf("%d%d", &n, &m);    for(int i = 1; i <= n; i ++)scanf("%d", &a[i]);    newnode(root, 0, -10000);    newnode(ch[root][1], root, -10000);    sz[root] ++;       build(ch[ch[root][1]][0], 1, n, ch[root][1]);    pushup(ch[root][1]); pushup(root);   // for(int i = 1; i <= 15; i ++)printf("%d %d %d  %d\n", i, num[i], lmax[i], maxx[i]);    char s[30]; int aa, bb, cc, dd, ee;      while(m --){scanf("%s", s); //  cout<<"!!!!"<<endl;    //  printf("!! %s", s);        if(s[0] == 'R'){            scanf("%d%d", &aa, &bb);            Reverse(aa, aa + bb - 1);        }        if(s[0] == 'G'){            scanf("%d%d", &aa, &bb);        //  cout<<"######"<<endl;            printf("%d\n", Get_sum(aa, aa + bb - 1));        }        if(s[0] == 'D'){            scanf("%d%d", &aa, &bb); n -= bb;            Delete(aa, aa + bb - 1);        }        if(s[0] == 'M' && s[2] == 'X'){        //  cout<<"!!!!"<<endl;            printf("%d\n", Max_sum());        }        if(s[0] == 'M' && s[2] == 'K'){            scanf("%d%d%d", &aa, &bb, &cc);            Make_same(aa, aa + bb - 1, cc);        }        if(s[0] == 'I'){            scanf("%d%d", &aa, &bb); n += bb;            for(int i = 1; i <= bb; i ++)scanf("%d", &a[i]);            Insert(aa, bb);        }        //  printf("%d\n", Get_sum(1, n));    /*      int pre; pre = 0;        for(int i = 1; i <= sz[root]; i ++){            splay(select(i), pre); pre = select(i);        }*///  for(int i = sz[root]; i >= 1; i --)printf("%d  %d %d   %d\n", maxx[select(i)], ch[select(i)][0], ch[select(i)][1], num[select(i)]);    }    return 0;}

bzoj 1208 宠物收养所

还是特别简单的一道裸题。

就是一个支持插入，删除，询问前驱和后继的splay。

我第一次在对 INF 操作的时候忘记了会爆int，所以WA 了一次，这种错误是在以后必须避免的。

12.11 : 我下面的这个代码虽然可以A但是是错的，应该再特判要删除的那个节点是不是在最开始建出来的虚点，因为这是早年的题所以数据比较水。。

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#define MAXN 80005#define INF (1<<31) - 1using namespace std;int n, ans, flag = -1, fat[MAXN], ch[MAXN][2], root, a[MAXN], cntt, now;inline void jie(int x, int y, int k){if(y)ch[y][k] = x; fat[x] = y;}inline void rotate(int x, int k){int y = fat[x];jie(ch[x][k], y, !k), jie(x, fat[y], ch[fat[y]][1] == y), jie(y, x, k);}void splay(int x, int goal){if(x == goal)return;while(fat[x] != goal){if(fat[fat[x]] == goal){rotate(x, ch[fat[x]][0] == x); break;}int y = fat[x], ky = ch[fat[y]][1] == y;if(ch[y][ky] == x)rotate(y, !ky), rotate(x, !ky);else rotate(x, ky), rotate(x, !ky);} if(goal == 0)root = x;}void newnode(int &x, int ff, int c){x = ++ cntt, a[cntt] = c, fat[cntt] = ff; }void insert(int x){int r = root;while(ch[r][a[r] < x])r = ch[r][a[r] < x];newnode(ch[r][a[r] < x], r, x);splay(cntt, 0);}int pre(int x){ int tmp = ch[x][0];if(!tmp)return 1;while(ch[tmp][1])tmp = ch[tmp][1]; return tmp;}int hou(int x){int tmp = ch[x][1];if(!tmp)return 2;while(ch[tmp][0])tmp = ch[tmp][0]; return tmp;}void del(int x){int pree = pre(x), houu = hou(x);splay(pree, 0), splay(houu, root);ch[houu][0] = 0;}void work(int x){insert(x); splay(cntt, 0);int pree = pre(cntt), houu = hou(cntt);int tmp = abs((long long)a[pree] - x) <= abs((long long)a[houu] - x) ? pree : houu;ans += abs(a[tmp] - x); ans %= 1000000;del(root);//printf("%d   %d\n", a[pree], a[houu]);splay(tmp, 0); del(tmp);}int main(){newnode(root, 0, INF + 1), newnode(ch[root][1], root, INF);scanf("%d", &n);while(n --){int k, c; scanf("%d%d", &k, &c);if(flag == -1 || flag == k)insert(c), now ++, flag = k;else work(c), now --;if(!now)flag = -1;}cout<<ans<<endl;return 0;}

bzoj 1507 editor

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