【ZOJ 2966】Build The Electric System（最小生成樹 Kruskal ）

Description

Input

Standard input will contain multiple test cases. The first line of the input is a single integer T (1 <= T <= 50) which is the number of test cases. And it will be followed by T consecutive test cases.

In each test case, the first line contains two positive integers N and E (2 <= N <= 500, N <= E <= N * (N - 1) / 2), representing the number of the villages and the number of the original power lines between villages. There follow E lines, and each of them contains three integers, A, B, K (0 <= A, B < N, 0 <= K < 1000). A and B respectively means the index of the starting village and ending village of the power line. If K is 0, it means this line still works fine after the snow storm. If K is a positive integer, it means this line will cost Kto reconstruct. There will be at most one line between any two villages, and there will not be any line from one village to itself.

Output

For each test case in the input, there's only one line that contains the minimum cost to recover the electric system to make sure that there's at least one way between every two villages.

Sample Input

1
3 3
0 1 5
0 2 0
1 2 9

Sample Output

5

題意

給出一個無向邊帶權圖，輸出最小生成樹

#include<iostream>
#include<algorithm>
using namespace std;
int p[10005],n,e;
struct edge
{
    int x,y,w;
}a[10005];
int cmp(edge a,edge b)
{
    return a.w<b.w;
}
int find(int r)
{
    if(p[r]!=r)
    p[r]=find(p[r]);
    return p[r];
}
int k()
{
    sort(a,a+e,cmp);              //一定要以邊edge排序啊！！！
    int ans=0,i;
    for(i=0;i<e;i++)
    {
        int fx=find(a[i].x),fy=find(a[i].y);
        if(fx!=fy)
        {
            p[fx]=fy;
            ans+=a[i].w;
        } 
    }
    return ans;
}
int main()  
{  
    int t,i;
    cin>>t;
    while(t--)
    {
        cin>>n>>e;
        for(i=0;i<=n;i++)
            p[i]=i;
        for(i=0;i<e;i++)
            scanf("%d%d%d",&a[i].x,&a[i].y,&a[i].w);
        printf("%d\n",k());
    } 
    return 0;
}