spoj 375 QTREE

题目链接：spoj 375

QTREE - Query on a tree

#tree

You are given a tree (an acyclic undirected connected graph) with N nodes, and edges numbered 1, 2, 3...N-1.

We will ask you to perfrom some instructions of the following form:

• CHANGE i ti : change the cost of the i-th edge to ti
  or
• QUERY a b : ask for the maximum edge cost on the path from node a to node b

Input

The first line of input contains an integer t, the number of test cases (t <= 20). t test cases follow.

For each test case:

• In the first line there is an integer N (N <= 10000),
• In the next N-1 lines, the i-th line describes the i-th edge: a line with three integers a b c denotes an edge between a, b of cost c (c <= 1000000),
• The next lines contain instructions "CHANGE i ti" or "QUERY a b",
• The end of each test case is signified by the string "DONE".

There is one blank line between successive tests.

Output

For each "QUERY" operation, write one integer representing its result.

Example

Input:131 2 12 3 2QUERY 1 2CHANGE 1 3QUERY 1 2DONEOutput:13

 Submit solution!

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题意：给出一棵树，每个边有一个边权，有2种操作，一种是改变第i个边的边权，另一种是查询从a到b的路径边权最大值。

题目分析：树链剖分的模板，2次dfs，之后映射到线段树上求区间最大值。

////  main.cpp//  spoj 375////  Created by teddywang on 2017/9/28.//  Copyright © 2017年 teddywang. All rights reserved.//#include <iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<vector>using namespace std;#define mst(a,b) memset(a,b,sizeof(a))#define lson l,mid,rt<<1#define rson mid+1,r,rt<<1|1const int maxn=11111;int T,n,m;int dep[maxn],size[maxn],fa[maxn],id[maxn],son[maxn],val[maxn],top[maxn];int num;vector<int> v[maxn];struct tree{    int x,y,val;    void read(){        scanf("%d%d%d",&x,&y,&val);    }}e[maxn];struct Tree{    int l,r,val;}tree[4*maxn];void dfs1(int u,int f,int d){    dep[u]=d;    size[u]=1;    son[u]=0;    fa[u]=f;    int len=v[u].size();    for(int i=0;i<len;i++)    {        int ff=v[u][i];        if(ff==f) continue;        else        {            dfs1(ff,u,d+1);            size[u]+=size[ff];            if(size[son[u]]<size[ff])            son[u]=ff;        }    }}void dfs2(int u,int tp){    top[u]=tp;    id[u]=++num;    if(son[u]) dfs2(son[u],tp);    int len=v[u].size();    for(int i=0;i<len;i++)    {        int ff=v[u][i];        if(ff==fa[u]||ff==son[u]) continue;        dfs2(ff,ff);    }}void pushup(int x){    tree[x].val=max(tree[x<<1].val,tree[x<<1|1].val);}void build(int l,int r,int rt){    tree[rt].l=l;    tree[rt].r=r;    if(l==r)    {        tree[rt].val=val[l];        return;    }    int mid=(l+r)>>1;    build(lson);    build(rson);    pushup(rt);}void update(int u,int v,int val){    if(tree[u].l==tree[u].r)    {        tree[u].val=val;        return;    }    int mid=(tree[u].l+tree[u].r)>>1;    if(v<=mid)        update(u<<1,v,val);    else update(u<<1|1,v,val);    pushup(u);}int query(int x,int l,int r){    if(tree[x].l>=l&&tree[x].r<=r)    {        return tree[x].val;    }    int mid=(tree[x].l+tree[x].r)>>1;    int ans=0;    if(l<=mid) ans=max(ans,query(x<<1,l,r));    if(r>mid) ans=max(ans,query(x<<1|1,l,r));    return ans;}int Yougth(int u, int v) {    int tp1 = top[u], tp2 = top[v];    int ans = 0;    while (tp1 != tp2) {        //printf("YES\n");        if (dep[tp1] < dep[tp2]) {            swap(tp1, tp2);            swap(u, v);        }        ans = max(query(1,id[tp1], id[u]), ans);        u = fa[tp1];        tp1 = top[u];    }    if (u == v) return ans;    if (dep[u] > dep[v]) swap(u, v);    ans = max(query(1,id[son[u]], id[v]), ans);    return ans;}int main(){    scanf("%d",&T);    while(T--)    {        scanf("%d",&n);        for(int i=1;i<n;i++)        {            e[i].read();            v[e[i].x].push_back(e[i].y);            v[e[i].y].push_back(e[i].x);        }        num=0;        dfs1(1,0,1);        dfs2(1,1);        for(int i=1;i<n;i++)        {            if(dep[e[i].x]<dep[e[i].y]) swap(e[i].x,e[i].y);            val[id[e[i].x]]=e[i].val;        }        build(1,num,1);        char s[20];        while(scanf("%s",s))        {            if(s[0]=='D') break;            int x,y;            scanf("%d%d",&x,&y);            if(s[0]=='Q')                printf("%d\n",Yougth(x,y));            else            {                update(1,id[e[x].x],y);            }        }        for(int i=0;i<=n;i++)            v[i].clear();    }}