hdu1879 继续畅通工程--prim

原题链接： ​​http://acm.hdu.edu.cn/showproblem.php?pid=1879​​

中文题目，不做解析了，就是一个简单的prim。


代码：

#define _CRT_SECURE_NO_DEPRECATE 

#include<iostream>
#include<cmath>
#include<algorithm>

using namespace std;

int n;
int p[105][105];
int lowCost[105];

int prim()
{
    int ans = 0;
    for (int i = 1; i <= n; i++)
        lowCost[i] = p[1][i];//默认1为起点
    lowCost[1] = -1;

    for (int i = 1; i < n; i++)
    {
        int min = 99999999;
        int minIndex;
        for (int j = 1; j <= n; j++)
        {
            if (lowCost[j] != -1 && lowCost[j] < min)
            {
                min = lowCost[j];
                minIndex = j;
            }
        }

        ans += min;
        lowCost[minIndex] = -1;

        for (int j = 1; j <= n; j++)
        {
            if (p[minIndex][j] < lowCost[j])
                lowCost[j] = p[minIndex][j];
        }
    }

    return ans;
}

int main()
{
    int a, b, cost, flag;
    while (~scanf("%d", &n) && n)
    {
        for (int i = 1; i <= n; i++)
            for (int j = 1; j <= 1; j++)
                p[i][j] = INT_MAX;

        for (int i = 1; i <= n*(n - 1) / 2; i++)
        {
            scanf("%d%d%d%d", &a, &b, &cost, &flag);
            if (flag == 1)
                p[a][b] = p[b][a] = 0;
            else
                p[a][b] = p[b][a] = cost;
        }

        printf("%d\n", prim());
    }
    return 0;
}