HDU 2586 最近公共祖先 LCA在线算法

How far away ？

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 13294    Accepted Submission(s): 4974

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

 

Source

ECJTU 2009 Spring Contest

 

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给一棵树，询问u，v之间的距离

#pragma comment(linker, "/STACK:1024000000,1024000000")#include <iostream>#include <string.h>#include <stdio.h>#include <vector>using namespace std;const int N = 50005;vector<int> v[N],w[N],query[N],num[N];int pre[N],dist[N],ans[N];bool vis[N];int n;void Init(){    for(int i=1; i<=n; i++)    {        v[i].clear();        w[i].clear();        query[i].clear();        num[i].clear();        pre[i] = i;        dist[i] = 0;        vis[i] = false;    }}int Find(int x){    if(pre[x] != x)        pre[x] = Find(pre[x]);    return pre[x];}void Union(int x,int y){    x = Find(x);    y = Find(y);    if(x == y) return;    pre[y] = x;}void Tarjan(int cur,int val){    vis[cur] = true;    dist[cur] = val;    int size = v[cur].size();    for(int i=0;i<size;i++)    {        int tmp = v[cur][i];        if(vis[tmp]) continue;        Tarjan(tmp,val + w[cur][i]);        Union(cur,tmp);    }    int Size = query[cur].size();    for(int i=0;i<Size;i++)    {        int tmp = query[cur][i];        if(!vis[tmp]) continue;        ans[num[cur][i]] = dist[cur] + dist[tmp] - 2*dist[Find(tmp)];    }}int main(){    int T,Q,x,y,z;    scanf("%d",&T);    while(T--)    {        scanf("%d%d",&n,&Q);        Init();        for(int i=1;i<n;i++)        {            scanf("%d%d%d",&x,&y,&z);            v[x].push_back(y);            w[x].push_back(z);            v[y].push_back(x);            w[y].push_back(z);        }        for(int i=0;i<Q;i++)        {            scanf("%d%d",&x,&y);            query[x].push_back(y);            query[y].push_back(x);            num[x].push_back(i);            num[y].push_back(i);        }        Tarjan(1,0);        for(int i=0;i<Q;i++)           printf("%d\n",ans[i]);    }    return 0;}