HDU

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 19232    Accepted Submission(s): 7525

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

 

Recommend

lcy

 

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题意：一个村庄有n个房子，有n-1条路径（类似一棵树），有q个查询，查询x到y的距离；

分析：树的节点之间的距离，LCA模板题

贴个模板：

#include <iostream>#include <algorithm>#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <vector>#include <queue>#include <stack>#include <string>#include <map>using namespace std;#define LL long long#define INF 0x3f3f3f3fconst int MAXN = 80080;const int MAXQ = 440;struct EdgeNode{int from;    int to;    int next;    int lca;}Edges[MAXN],QEdges[MAXQ];int Head[MAXN],QHead[MAXN];//离线查询int father[MAXN],Dist[MAXN];bool vis[MAXN];int id,ip;int Find(int x){    return father[x]==x?x:father[x]=Find(father[x]);}void LCA(int u){    father[u] = u;    vis[u] = true;    for(int k = Head[u]; k != -1; k = Edges[k].next)    {        if(!vis[Edges[k].to])        {            Dist[Edges[k].to] = Dist[u] + Edges[k].lca;            LCA(Edges[k].to);            father[Edges[k].to] = u;        }    }    for(int k = QHead[u]; k != -1; k = QEdges[k].next)    {        if(vis[QEdges[k].to])        {            QEdges[k].lca = Dist[u] + Dist[QEdges[k].to] - 2*Dist[Find(QEdges[k].to)];            QEdges[k^1].lca = QEdges[k].lca;        }    }}void init(){id=ip=0;memset(father,0,sizeof(father));    memset(Dist,0,sizeof(Dist));    memset(vis,false,sizeof(vis));    memset(Head,-1,sizeof(Head));    memset(QHead,-1,sizeof(QHead));}void addedge(int u,int v,int w){Edges[id].from=u;Edges[id].to = v;    Edges[id].lca = w;    Edges[id].next = Head[u];    Head[u] = id++;}void addQedge(int u,int v){QEdges[ip].from = u;QEdges[ip].to = v;    QEdges[ip].next = QHead[u];    QHead[u] = ip++;}int main(){    int t;    scanf("%d",&t);    while(t--)    {        int n,m;        scanf("%d%d",&n,&m);        init();        for(int i = 2; i <= n; ++i)        {            int u,v,w;            scanf("%d%d%d",&u,&v,&w);            addedge(u,v,w);            addedge(v,u,w);        }        for(int i = 1; i <= m; ++i)        {        int u,v;            scanf("%d%d",&u,&v);            addQedge(u,v);            addQedge(v,u);        }        LCA(1);        for(int i = 0; i < ip; i += 2)            printf("%d\n",QEdges[i].lca);    }    return 0;}