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题目链接:

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

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

2

3 2

1 2 10

3 1 15

1 2

2 3

2 2

1 2 100

1 2

2 1

Sample Output

10

25

100

100

Problem solving report:

Description: 在一个村庄里有n个地方，这n个地方每两个之间都有一条唯一的双向道。进行m次询问，A到B的距离是多少？

Problem solving: 利用vector,把每条路的起点当数组的下标，终点作为数组里面存的内容，然后把权值用另一个数组存一下(定义一个结构体)，在深搜里面就把每个数组里面存的终点作为新的起点进入新的递归。直到找到终点。

#include 
   
     //动态数组的头文件#include 
    
     #include 
     
      #define N 40010using namespace std;struct edge {    int v, w;};int vis[N], temp;vector 
      
        a[N]; //定义一个结构体动态二维数组void DFS(int x, int y, int sum){    int i;    if (x == y)    {        printf("%d\n", sum);        temp = 1;        return ;    }    int len = a[x].size(); //返回数组a[x]的元素个数    if (!len || vis[x] || temp)        return ;    for (int i = 0; i < len; i++)    {        vis[x] = 1;        DFS(a[x][i].v, y, sum + a[x][i].w);        vis[x] = 0;    }}int main(){    int t, m, n, u, v, w, x, y;    scanf("%d", &t);    while (t--)    {        scanf("%d%d", &n, &m);        memset(vis, 0, sizeof(vis));        for (int i = 0; i < n; i++)            a[i].clear(); //清空动态数组         for (int i = 1; i < n; i++)        {            scanf("%d%d%d", &u, &v, &w);            a[u].push_back((edge){v, w});            a[v].push_back((edge){u, w});        }        for (int i = 0; i < m; i++)        {            temp = 0;            scanf("%d%d", &x, &y);            DFS(x, y, 0);        }    }    return 0;}
      
     
    
   

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