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

Line 1: The number of farms, N (3 <= N <= 100).
Line 2: The subsequent lines contain the N x N connectivity 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.

Sample Input

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

Sample Ouput

28

純模板題，最小生成樹prim之鄰接矩陣實現，而這個模板我覺得特別漂亮，很簡潔，也很清晰。

借鑑了下面這個部落格，感謝一下。
TRTTG

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>

using namespace std;
const int N=1000+10;
const int INF=0x3f3f3f3f;
int n,ans=0,tmp,k;
int dist[N],vis[N],a[N][N];

int main(){
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);

    cin>>n;
    for(int i=1;i<=n;i++){
        for(int j=1;j<=n;j++){
            cin>>a[i][j];
        }
    }

    memset(dist,0x3f,sizeof(dist));
    dist[1]=0;
    for(int i=1;i<=n;i++){
        tmp=INF;
        for(int j=1;j<=n;j++){
            if (!vis[j]&&dist[j]<tmp){
                tmp=dist[j];
                k=j;
            }
        }
        ans+=dist[k];
        vis[k]=1;
        for(int j=1;j<=n;j++){
            if (!vis[j]&&a[k][j]<dist[j]) dist[j]=a[k][j];
        }
    }   

    cout<<ans<<endl;
    return 0;
}