POJ 1258 Agri-Net （Prim普里姆算法应用）

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 23658		Accepted: 9306

Description

Farmer John has been elected mayor of his town! One of his campaign promises was to bring internet connectivity to all farms in the area. He needs your help, of course.
Farmer John ordered a high speed connection for his farm and is going to share his connectivity with the other farmers. To minimize cost, he wants to lay the minimum amount of optical fiber to connect his farm to all the other farms.
Given a list of how much fiber it takes to connect each pair of farms, you must find the minimum amount of fiber needed to connect them all together. Each farm must connect to some other farm such that a packet can flow from any one farm to any other farm.
The distance between any two farms will not exceed 100,000.

Input

The input includes several cases. For each case, the first line contains the number of farms, N (3 <= N <= 100). The following lines contain the N x N conectivity matrix, where each element shows the distance from on farm to another. Logically, they are N lines of N space-separated integers. Physically, they are limited in length to 80 characters, so some lines continue onto others. Of course, the diagonal will be 0, since the distance from farm i to itself is not interesting for this problem.

Output

For each case, output a single integer length that is the sum of the minimum length of fiber required to connect the entire set of farms.

Sample Input

4
0 4 9 21
4 0 8 17
9 8 0 16
21 17 16 0

Sample Output

28

Source

USACO 102
//普里姆算法
根据王晓东的算法分析与设计：
设G = （v， E）是连通的带权图，V={1，2，3，…，n}，构造最小生成树的prim的算法：初始化S={1}，只要S是V的真子集，那
么就可以用贪心策略来找最小边，选取在S中的i，以及在V-S中的j，且c［i］［j］最小边，将顶点j添加到S中。这个过程直到S == V
为止。

#include <iostream>
#include <cstdio>
using namespace std;

#define N 101
#define INF 1<<30

int c[N][N], lowcost[N], closest[N], visited[N];

void prim(int n)
{
    int i, j, k, min, total = 0;
    for(i = 1; i <= n; i++)
    {
        lowcost[i] = c[1][i];
        visited[i] = false;
        closest[i] = 1;
    }
    visited[1] = true;
    for(i = 2; i <= n; i++)
    {
        k = -1;
        min = INF;
        for(j = 2; j <= n; j++)
        {
            if((!visited[j]) && (lowcost[j] < min))  //找出与S中点有关联的最小边
            {
                min = lowcost[j];
                k = j;
            }
        }
        if(k == -1) break;
        total += min;
        visited[k] = true;
        for(j = 2; j <= n; j++)
        {

            if((!visited[j]) && (lowcost[j] > c[k][j]))  //再将与 k点有关的且在V-S中的点联起来的边更新到lowcost中    

            {

                lowcost[j] = c[k][j];

                closest[j] = k;
            }
        }
    }
    printf("%d\n", total);
}

int main()
{
    int i, j, n;
    while(scanf("%d", &n) != EOF)
    {
        for(i = 1; i <= n; i++)
        {
            for(j = 1; j<= n; j++)
                scanf("%d", &c[i][j]);
        }
        prim(n);
    }
    return 0;
}