POJ 2763 Housewife Wind 树链剖分

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题目大意：

就是给出一棵树然后p次操作, 每次操作询问从当前位置到某个位置的路径的权值和, 另外一个操作是修改某条边的权值

大致思路：

树链剖分第二题...

因为写线段树的在建树的时候没有pushUp调试了一个小时才发现....感觉自己智商好捉急.....

不过因为这个原因对于树链剖分的过程更熟悉了...

首先在处理边的时候, 我们先将边上的信息转移到点上, 这里我们将所有的边的信息转移到其深度更大的结点上

于是在处理两点之间的距离时, 沿着剖分之后的路径走的时候, 如果到了top[x] == top[y]的情况(top[x] 表示点x所在链的深度最小的点, 即链的顶点)

如果x == y那么就直接结束, 如果top[x] < top[y]则应该加上x到hson[y]之间的点的权值和, hson[y]为y的重儿子(top[x] == top[y] && x != y只会出现在重链上)

于是整个过程就很简单了, 整体复杂度O(n*logn*logn)

代码如下：

Result : Accepted Memory : 8912 KB Time : 1610 ms

/* * Author: Gatevin * Created Time:  2015/9/7 20:41:59 * File Name: Sakura_Chiyo.cpp */#include<iostream>#include<sstream>#include<fstream>#include<vector>#include<list>#include<deque>#include<queue>#include<stack>#include<map>#include<set>#include<bitset>#include<algorithm>#include<cstdio>#include<cstdlib>#include<cstring>#include<cctype>#include<cmath>#include<ctime>#include<iomanip>using namespace std;const double eps(1e-8);typedef long long lint;#define maxn 100010int top[maxn];int grandson[maxn];int dep[maxn];int siz[maxn];int belong[maxn];int father[maxn];int Q[maxn];int cnt;int hson[maxn];int n, q, s;bool vis[maxn];int id[maxn];int antiID[maxn];struct Edge{    int u, v, w, nex;    Edge(int _u, int _v, int _w, int _nex)    {        u = _u, v = _v, w = _w, nex = _nex;    }    Edge(){}};int head[maxn];//用于邻接表int tot;//总边数Edge edge[maxn << 1];int w[maxn];//将其父边的权值信息降到这个结点处void add_Edge(int x, int y, int w){    edge[++tot] = Edge(x, y, w, head[x]);    head[x] = tot;    return;}void split(){    cnt = 0;    int l = 0, r = 1;    dep[Q[r] = 1] = 1;    father[r] = -1;    w[r] = 0;    while(l < r)    {        int x = Q[++l];        if(head[x] == -1) continue;        for(int j = head[x]; j + 1; j = edge[j].nex)        {            int y = edge[j].v;            if(y == father[x]) continue;            w[y] = edge[j].w;            dep[Q[++r] = y] = dep[x] + 1;            father[y] = x;        }    }    for(int i = n; i; i--)    {        int x = Q[i], p = -1;        siz[x] = 1;        if(head[x] == -1) continue;        for(int j = head[x]; j + 1; j = edge[j].nex)        {            int y = edge[j].v;            if(y == father[x]) continue;            siz[x] += siz[y];            if(p == -1 || (p > 0 && siz[y] > siz[p]))                p = y;        }        if(p == -1)        {            hson[x] = -1;            grandson[++cnt] = x;            belong[top[cnt] = x] = cnt;        }        else        {            hson[x] = p;            belong[x] = belong[p];            top[belong[x]] = x;        }    }    int idx = 0;    memset(vis, 0, sizeof(vis));    for(int i = n; i; i--)//完成树上的点到线段树控制的区间的映射    {        int x = Q[i];        if(vis[x]) continue;        vis[x] = 1;        id[x] = ++idx;//实际上的结点x对应的区间位置        antiID[idx] = x;        while(father[x] != -1 && belong[father[x]] == belong[x] && !vis[father[x]])        {            x = father[x];            id[x] = ++idx;            antiID[idx] = x;            vis[x] = 1;        }    }    return;}//线段树单点修改区间求和#define lson l, mid, rt << 1#define rson mid + 1, r, rt << 1 | 1int val[maxn << 2];void pushUp(int rt){    val[rt] = val[rt << 1] + val[rt << 1 | 1];    return;}void build(int l, int r, int rt){    if(l == r)    {        val[rt] = w[antiID[l]];        return;    }    int mid = (l + r) >> 1;    build(lson);    build(rson);    pushUp(rt);    return;}void update(int l, int r, int rt, int pos, int value){    if(l == r)    {        val[rt] = value;        return;    }    int mid = (l + r) >> 1;    if(pos <= mid) update(lson, pos, value);    else update(rson, pos, value);    pushUp(rt);    return;}int query(int l, int r, int rt, int L, int R){    if(l >= L && r <= R)        return val[rt];    int mid = (l + r) >> 1;    int ret = 0;    if(mid >= L) ret += query(lson, L, R);    if(mid + 1 <= R) ret += query(rson, L, R);    return ret;}int answer(int x, int y)//询问x到y的路径和{    int ans = 0;    while(top[belong[x]] != top[belong[y]])    {        if(dep[top[belong[x]]] < dep[top[belong[y]]])            swap(x, y);        ans += query(1, n, 1, id[x], id[top[belong[x]]]);        x = father[top[belong[x]]];    }    if(x == y) return ans;//注意这里结点代表的是向上与父亲节点连边的权值, 当x == y的时候直接结束    //否则两者在同一条链上, 且一定是重链    if(dep[x] < dep[y]) swap(x, y);//x是在下面的那个    ans += query(1, n, 1, id[x], id[hson[y]]);//hson[y]是y的重儿子(考虑到点权代表的是与父亲连边的边权)    return ans;}void change(int x, int w){    x <<= 1;//双向边    int u = edge[x].u, v = edge[x].v;    if(father[u] == v)        update(1, n, 1, id[u], w);    else update(1, n, 1, id[v], w);    return;}int main(){    while(scanf("%d %d %d", &n, &q, &s) != EOF)    {        memset(head, -1, sizeof(head));        tot = 0;        int u, v, w;        for(int i = 1; i < n; i++)        {            scanf("%d %d %d", &u, &v, &w);            add_Edge(u, v, w);            add_Edge(v, u, w);        }        split();        build(1, n, 1);        int op, x;        while(q--)        {            scanf("%d", &op);            switch(op)            {                case 0: scanf("%d", &u);                        printf("%d\n", answer(s, u));                        s = u;                        break;                case 1: scanf("%d %d", &x, &w);//将第x条边的权值改为w                        change(x, w);                        break;            }        }    }    return 0;}/*11 10 11 2 12 6 22 7 37 8 47 11 51 3 93 5 53 9 79 10 101 4 80 80 110 50 50 100 30 61 3 1990 110 4*/

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