Agri-Net Time Limit:1000MS     Memory Limit:10000KB  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

題意：求連通所有村莊的最短距離和。

思路：prim演算法。

初始集合只有一個村莊1，然後迴圈選擇dist[x]（初始化=v[x][1]）最小的放入集合，sum+=dist[x];

use[x]=1表示x村莊已在集合的連通村莊中。

然後從1~n更新集合外村莊的dist[i]=dist[i]<v[i][x]?dist[i]:v[i][x];

迴圈直至集合外村莊為0；

#include<iostream>
#include<stdio.h>
#include<vector>
#include<string.h>
#include<algorithm>
using namespace std;
int dist[200],v[200][200],use[200];
bool cmp(const int &a,const int &b)
{
	return dist[a]<dist[b];
}
int main()
{
	int n,i,j;
	int sum=0;
	int a;
	while(scanf("%d",&n)!=EOF)
	{
		sum=0;
		memset(use,0,sizeof(use));
		vector<int> vec;
		for(i=1;i<=n;i++)
		{
			for(j=1;j<=n;j++)
				scanf("%d",&v[i][j]);
			dist[i]=v[i][1];
			if(i!=1) vec.push_back(i);//儲存集合外村莊
		}
		use[1]=1;
		sort(vec.begin(),vec.end(),cmp);
		while(!vec.empty())
		{
			a=vec[0];
			vec.erase(vec.begin());//刪除 表示已放入集合
		    sum+=dist[a];
		    use[a]=1;
			for(i=1;i<=n;i++)
				if(!use[i]) dist[i]=dist[i]<v[i][a]?dist[i]:v[i][a];
				sort(vec.begin(),vec.end(),cmp);//將更新dist後的村莊按從小到大排序
		}
		printf("%d\n",sum);
	}
	return 0;
}