zoj2617Edison(splay模拟)

题目请戳这里

题目大意：模拟一个简单的洗牌过程。C张牌，编号0~c-1，一次洗牌就是将从第p张牌开始的连续l张整体移到最前面。现在给s个操作，每个操作有r次重复洗牌动作，求洗完牌后的c张牌序列中，奇数位置的牌点数之和。

题目分析：由于只有一个操作，所以直接模拟一下就可以了。洗牌过程看起来很复杂，看穿了也就没什么了。一次操作相当于将前p张牌循环右移l的位置，重复r次，其实就是将前p张牌循环右移l*r次。直接模拟就可以了。

兴高采烈的撸了个splay，好慢的样子。。。早知道就直接用愉快的数组模拟了。

详情请见代码：

#include <iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int N = 1000005;const int inf = 0x3f3f3f3f;int next[N];struct node{    int l,r,f,lazy,sizea,key;}tree[N];int n,m,cnt,c,s;long long ans;void init(){    tree[0].sizea = tree[0].lazy = tree[0].l = tree[0].r = tree[0].f = 0;    tree[0].key = inf;    for(int i = 0;i < N;i ++)        next[i] = i + 1;}int newnode(){    int p = next[0];    next[0] = next[p];    tree[p].lazy = 0;    tree[p].f = tree[p].l = tree[p].r = 0;    tree[p].sizea = 1;    return p;}void delnode(int p){    next[p] = next[0];    next[0] = p;}void pushup(int rt){    if(rt)        tree[rt].sizea = tree[tree[rt].l].sizea + tree[tree[rt].r].sizea + 1;}void pushdown(int rt){    if(!rt)        return;    int ls = tree[rt].l;    int rs = tree[rt].r;    if(tree[rt].lazy)    {        swap(tree[rt].l,tree[rt].r);        if(ls)            tree[ls].lazy ^= 1;        if(rs)            tree[rs].lazy ^= 1;        tree[rt].lazy = 0;    }}void zig(int x){    int p = tree[x].f;    pushdown(p);    pushdown(x);    tree[p].l = tree[x].r;    if(tree[x].r)        tree[tree[x].r].f = p;    pushup(p);    tree[x].r = p;    tree[x].f = tree[p].f;    pushup(x);    tree[p].f = x;    if(tree[x].f == 0)        return;    if(tree[tree[x].f].l == tree[x].r)        tree[tree[x].f].l = x;    else        tree[tree[x].f].r = x;}void zag(int x){    int p = tree[x].f;    pushdown(p);    pushdown(x);    tree[p].r = tree[x].l;    if(tree[x].l)        tree[tree[x].l].f = p;    pushup(p);    tree[x].l = p;    tree[x].f = tree[p].f;    pushup(x);    tree[p].f = x;    if(tree[x].f == 0)        return;    if(tree[tree[x].f].l == tree[x].l)        tree[tree[x].f].l = x;    else        tree[tree[x].f].r = x;}int splay(int x,int goal){    pushdown(x);    while(tree[x].f != goal)    {        int p = tree[x].f;        int g = tree[p].f;        if(g == goal)        {            if(tree[p].l == x)                zig(x);            if(tree[p].r == x)                zag(x);        }        else        {            if(tree[g].l == p && tree[p].l == x)                zig(p),zig(x);            else if(tree[g].l == p && tree[p].r == x)                zag(x),zig(x);            else if(tree[g].r == p && tree[p].r == x)                zag(p),zag(x);            else if(tree[g].r == p && tree[p].l == x)                zig(x),zag(x);        }    }    pushup(x);    return x;}int build(int l,int r,int fa){    if(l > r)        return 0;    int mid = (l + r)>>1;    int p = newnode();    tree[p].l = build(l,mid - 1,p);    tree[p].key = cnt ++;    tree[p].f = fa;    tree[p].r = build(mid + 1,r,p);    pushup(p);    return p;}void prepare(int &root){    root = newnode();    tree[root].key = inf;    tree[root].r = newnode();    tree[tree[root].r].key = inf;    tree[tree[root].r].f = root;    tree[tree[root].r].l = build(1,c,tree[root].r);    pushup(tree[root].r);    pushup(root);}int finda(int pos,int rt){    if(!rt)        return 0;    pushdown(rt);    if(tree[tree[rt].l].sizea == pos - 1)        return rt;    if(tree[tree[rt].l].sizea >= pos)        return finda(pos,tree[rt].l);    else        return finda(pos - tree[tree[rt].l].sizea - 1,tree[rt].r);}void Rotate_interval(int a,int b,int &root){    int l = finda(a,root);    int r = finda(b + 2,root);    root = splay(l,0);    tree[root].r = splay(r,root);    pushup(tree[root].r);    pushup(root);}void Reverse(int a,int b,int &root){    Rotate_interval(a,b,root);    tree[tree[tree[root].r].l].lazy ^= 1;}void dfs(int rt){    if(!rt)        return;    pushdown(rt);    dfs(tree[rt].l);    if(tree[rt].key != inf)        cnt ++;    if(cnt&1)        ans += tree[rt].key;    dfs(tree[rt].r);    delnode(rt);}int main(){    int t,i,cas,root;    int pos,len,rep;    cas = 0;    init();    scanf("%d",&t);    while(t --)    {        scanf("%d%d",&c,&s);        ans = 0;        cnt = 0;        prepare(root);        while(s --)        {            scanf("%d%d%d",&pos,&len,&rep);            int length = len * rep;            len += pos;            length %= len;            if(length == 0)                continue;            length = len - length;            Reverse(1,length,root);            Reverse(length + 1,len,root);            Reverse(1,len,root);//debug(root        }        cnt = 0;        dfs(root);        printf("Case %d:\n",++cas);        printf("%lld\n",ans);        if(t)            puts("");    }    return 0;}

补上数组模拟的代码：

so easy!so fast!!!

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int N = 1000005;int lcm[N],temp[N];long long ans;int c,s,p,l,r;int main(){    int t,i,j;    int cas = 0;    scanf("%d",&t);    while(t --)    {        scanf("%d%d",&c,&s);        for(i = 1;i <= c;i ++)            lcm[i] = i - 1;        while(s --)        {            scanf("%d%d%d",&p,&l,&r);            int len = l * r;            l += p;            len %= l;            if(len == 0)                continue;            if(len >= (l>>1))            {                for(i = 1;i <= l - len;i ++)                    temp[i] = lcm[i];                for(;i <= l;i ++)                    lcm[i - l + len] = lcm[i];                for(i = len + 1;i <= l;i ++)                    lcm[i] = temp[i - len];            }            else            {                for(i = l,j = 1;i >= l - len + 1;i --)                    temp[j ++] = lcm[i];                for(;i >= 1;i --)                    lcm[i + len] = lcm[i];                for(i = j - 1;i >= 1;i --)                    lcm[len - i + 1] = temp[i];            }        }        ans = 0;        for(i = 1;i <= c;i += 2)            ans += lcm[i];        printf("Case %d:\n",++cas);        printf("%lld\n",ans);        if(t)            puts("");    }    return 0;}