How far away ？ HDU

How far away ？

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 17281    Accepted Submission(s): 6662

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

题意：求最短路。

解题思路：使用两个向量数组，一个存边关系，一个存权值，按照先后顺序，找最短路时，再使用站的先入先出的思想，向四周发散开来，避免成圈，需要进行标记，再用一个数组存，从初始点到每个点的距离。代码很好理解，如果不理解怎样存的权值，可以动手模拟一下，就理解了。

#include <stdio.h>#include <string.h>#include <algorithm>#include <vector>#include <stack>using namespace  std;#define maxn 40000+10vector<int>V[maxn];vector<int>W[maxn];int n,m,vis[maxn],dis[maxn];void Inf(){    for(int i=1; i<=n; i++)    {        V[i].clear();        W[i].clear();    }}int  dfs(int x,int y){    stack<int>Q;    memset(vis,0,sizeof(vis));    memset(dis,0,sizeof(dis));    Q.push(x);    vis[x]=1;    while(!Q.empty())    {        int p=Q.top();        Q.pop();        if(p==y)            break;        for(int i=0; i<V[p].size(); i++)        {            int t=V[p][i];            if(vis[t]) continue;            vis[t]=1;            Q.push(t);            dis[t]=dis[p]+W[p][i];      //W向量，存的权值        }    }    return dis[y];}int main(){    int t;    scanf("%d",&t);    while(t--)    {        Inf();        scanf("%d %d",&n,&m);        int a,b,c;        for(int i=1; i<n; i++)        {            scanf("%d %d %d",&a,&b,&c);            V[a].push_back(b);       //无向图，双向存储            V[b].push_back(a);            W[a].push_back(c);              W[b].push_back(c);      //方便，后面的权值向加。        }        for(int i=1; i<=m; i++)        {            scanf("%d %d",&a,&b);            printf("%d\n",dfs(a,b));        }    }    return 0;}