BNU 16485 Build The Electric System

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本文详细介绍了在遭受严重雪灾导致电力系统损坏的情况下，如何通过编程算法计算出重建电力线路以确保所有村庄间至少有一条连通路径所需的最低成本。通过案例分析和具体步骤阐述，旨在为专业程序员提供解决实际问题的策略。

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Build The Electric System

Time Limit: 2000ms

Memory Limit: 65536KB

This problem will be judged on ZJU. Original ID: 2966
64-bit integer IO format: %lld Java class name: Main

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In last winter, there was a big snow storm in South China. The electric system was damaged seriously. Lots of power lines were broken and lots of villages lost contact with the main power grid. The government wants to reconstruct the electric system as soon as possible. So, as a professional programmer, you are asked to write a program to calculate the minimum cost to reconstruct the power lines to make sure there's at least one way between every two villages.

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 andB 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 K to 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

Sample Output

Source

The 5th Zhejiang Provincial Collegiate Programming Contest

Author

ZHOU, Ran

题意：n个村庄 e条线路 k是线路的修理费用问最少要多少钱才能恢复这些村庄的供电

思路：先对e条线进行排序把修理费用少的放在前面

然后用并查一下看哪几个是联通的就不用修了

再把不同的修好联通起来就ok了

#include<stdio.h>
#include<algorithm>
using namespace std;
int father[50000];
int find(int x)
{
    return father[x]==x?father[x]:find(father[x]);
}
struct ml
{
    int a,b,k;
}p[50000];
int cmp(ml x,ml y)
{
    return x.k<y.k;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n,e;
        scanf("%d%d",&n,&e);

        for(int i=0;i<e;i++)
            scanf("%d%d%d",&p[i].a,&p[i].b,&p[i].k);
        sort(p,p+e,cmp);

        for(int i=0;i<n;i++)
            father[i]=i;
        int sum=0;

        for(int i=0;i<e;i++)
        {
            int fx=find(p[i].a);
            int fy=find(p[i].b);
            if(fx!=fy)
            {
                father[fx]=fy;
                sum+=p[i].k;
            }
        }
        printf("%d\n",sum);
    }
    return 0;
}