Pku oj 1528 Agri-Net(MST）

                                                                                                           Agri-Net

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 49344 Accepted: 20494

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

USACO 102

给你一个邻接矩阵，求MST，果断Prim

#include<cstdio>#include<iostream>#include<algorithm>#include<cstring>#include<cstdlib>#include<cmath>#include<queue>using namespace std;const int maxn = 305;const int Inf = 0x3f3f3f;int dis[maxn];int map[maxn][maxn];int vis[maxn];int N;void init(){    for(int i=0;i<N;i++)    {        dis[i] = map[i][0];        vis[i] = 0;    }} //距离和访问数组初始化int prim(){    init();    dis[0] = 0; //起点距离初始化为0    vis[0] = 1; //标记起点    int ans = 0; //权值和初始化为0    for(int i=0;i<N-1;i++)    {        int temp = Inf; // 记录最小边        int flag; //纪录最小点        for(int j=0;j<N;j++)        {            if(!vis[j] && dis[j] < temp)            {                temp = dis[j];                flag = j;            }        } //找最小边        vis[flag] = 1; //标记选出的边        ans += dis[flag]; //把边加入生成树中        for(int j=0;j<N;j++)        {            if(!vis[j] && dis[j] > map[flag][j])                dis[j] = map[flag][j]; //更新最小权值        }    }    return ans;}int main(){    while(~scanf("%d",&N))    {        for(int i=0;i<N;i++)        {            for(int j=0;j<N;j++)                scanf("%d",&map[i][j]); //邻接矩阵存图        }        int ans = prim();        printf("%d\n",ans);    }    return 0;}