poj 1258 Agri-Net（最小生成树 prime算法）

Agri-Net
http://poj.org/problem?id=1258

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 40013 Accepted: 16283

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

40 4 9 214 0 8 179 8 0 1621 17 16 0

Sample Output

28

Source

题目：

第一行，输入一个数n；

接下来n*n的矩阵，map[i][j代表i，j之间边的权值；

输出：把所有点连接起来的最小权值；

prime 算法：把顶点看成两个集合，一个叫做确定的集合a，一个叫做待确定的集合b；

然后开始时所有的顶点均在b中，从b中任意找出一个顶点，然后将其visit数组标为真，找出其与邻接顶点c的权值最小的那条边，ans累加；

然后再从c开始，继续上述步骤，知道找到了n-1条边跳出循环；输出累加结果ans；

代码：

#include <iostream>#include <stdio.h>#include <string.h>#include<stdlib.h>#define MAX 110#define INF 1e9using namespace std;int map[MAX][MAX];int visit[MAX];int low[MAX],n;int pos;int min(int i){    int j;    int ans=INF;    for(j=1; j<=n; j++)    {        if(i!=j&&!visit[j])        {            low[j]=min(low[j],map[i][j]);//更新low数组            if(ans>low[j])//记录下更新完后low数组的最小值            {                ans=low[j];                pos=j;            }        }    }    return ans;}int main(){    int i,j,b,ans,cnt;    while(~scanf("%d",&n))    {        cnt=0;        memset(visit,0,sizeof(visit));        memset(low,1,sizeof(low));        memset(map,0,sizeof(map[0])*MAX);        for(i=1; i<=n; i++)        {            for(j=1; j<=n; j++)            {                scanf("%d",&b);                map[i][j]=b;//将图存入邻接矩阵中            }        }        ans=0;        for(i=1,pos=i; i<=n; i=pos)        {            if(!visit[i])            {                cnt++;                ans+=min(i);//求low数组的最小值并累加权值                visit[i]=1;                if(cnt==n-1)//找到n-1条边后跳出                    break;            }        }        printf("%d\n",ans);    }    return 0;}