HDU

How far away ？

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 10580    Accepted Submission(s): 3851

Problem Description

There are n houses in the village and some bidirectional roads connecting them. Every day peole always like to ask like this "How far is it if I want to go from house A to house B"? Usually it hard to answer. But luckily int this village the answer is always unique, since the roads are built in the way that there is a unique simple path("simple" means you can't visit a place twice) between every two houses. Yout task is to answer all these curious people.

 

Input

First line is a single integer T(T<=10), indicating the number of test cases.
  For each test case,in the first line there are two numbers n(2<=n<=40000) and m (1<=m<=200),the number of houses and the number of queries. The following n-1 lines each consisting three numbers i,j,k, separated bu a single space, meaning that there is a road connecting house i and house j,with length k(0<k<=40000).The houses are labeled from 1 to n.
  Next m lines each has distinct integers i and j, you areato answer the distance between house i and house j.

 

Output

For each test case,output m lines. Each line represents the answer of the query. Output a bland line after each test case.

 

Sample Input

23 21 2 103 1 151 22 32 21 2 1001 22 1

 

Sample Output

1025100100

题意：给出一棵树，问树上两点距离

#include <bits/stdc++.h>using namespace std;const int N = 4e4 + 10;int t, n, m;int tot;struct xx{    int to, val, nex;} a[N<<1];int dep[N], dis[N];int fa[N][20], head[N];void Init(){    tot = 0;    memset(a, 0, sizeof a);    memset(dis, 0, sizeof dis);    memset(dep, 0, sizeof dep);    memset(fa, 0, sizeof fa);    memset(head, -1, sizeof head);}void Add(int u, int v, int w){    a[++tot].to = v, a[tot].nex = head[u];    a[tot].val = w, head[u] = tot;}void dfs(int p, int f, int step, int deep){    dis[p] = step, dep[p] = deep;    fa[p][0] = f;    for(int i = 1; i <= log2(dep[p]); i++){        fa[p][i] = fa[fa[p][i-1]][i-1];    }    for(int i = head[p]; i != -1; i = a[i].nex){        if(a[i].to != f){            dfs(a[i].to, p, step+a[i].val, deep+1);        }    }}int Lca(int u, int v){    if(dep[u] < dep[v]) swap(u, v);    for(int i = 20; i >= 0; i--){        if(dep[u] == dep[v]) break;        if(dep[fa[u][i]] < dep[v]) continue;        u = fa[u][i];    }    if(u == v) return u;    for(int i = log2(dep[u]); i >= 0; i--){        if(fa[u][i] == fa[v][i]) continue;        u = fa[u][i], v = fa[v][i];    }    return fa[u][0];}int main(){    scanf("%d", &t);    while(t--){        scanf("%d%d", &n, &m);        Init();        int x, y, z;        for(int i = 1; i < n; i++){            scanf("%d%d%d", &x, &y, &z);            Add(x, y, z); Add(y, x, z);        }        dfs(1, 0, 0, 0);        while(m--){            scanf("%d%d", &x, &y);            printf("%d\n", dis[x]+dis[y]-2*dis[Lca(x, y)]);        }    }}