POJ - 2421 Constructing Roads （最小生成树）

There are N villages, which are numbered from 1 to N, and you should build some roads such that every two villages can connect to each other. We say two village A and B are connected, if and only if there is a road between A and B, or there exists a village C such that there is a road between A and C, and C and B are connected.

We know that there are already some roads between some villages and your job is the build some roads such that all the villages are connect and the length of all the roads built is minimum.
Input
The first line is an integer N (3 <= N <= 100), which is the number of villages. Then come N lines, the i-th of which contains N integers, and the j-th of these N integers is the distance (the distance should be an integer within [1, 1000]) between village i and village j.

Then there is an integer Q (0 <= Q <= N * (N + 1) / 2). Then come Q lines, each line contains two integers a and b (1 <= a < b <= N), which means the road between village a and village b has been built.
Output
You should output a line contains an integer, which is the length of all the roads to be built such that all the villages are connected, and this value is minimum.
Sample Input
3
0 990 692
990 0 179
692 179 0
1
1 2
Sample Output
179
问题链接：http://poj.org/problem?id=2421
问题简述：输入n代表有n座城市，然后以下n行，第i行有n个数表示第i个城市到每个城市的距离，还有已经连接了的城市。求能链接所有城市的最短距离
问题分析：最小生成树…先把所有点的距离设置为最大，把已经连接的城市距离设置为0。从第一个点开始，找出与第一个点最短距离的点，标记访问过，再找出一，二号点所有的最短距离，然后以此类推，找出与前n个已经连接的点的最短距离，直到将所有点连接，把所有距离累加输出即可
AC通过的C++语言程序如下：

#include
#include 
#include
#include
#include
#include 
#include
#include
#include
using namespace std;
const int inf = 999999999;
const int N = 105;
int map[N][N], dis[N], n, m;
bool vis[N];
int prim()
{
	for (int i = 1; i <= n; i++)
	{
		dis[i] = inf;
	}
	dis[1] = 0;
	for (int i = 1; i <= n; i++)
	{
		int t = inf;
		int a;
		for (int j = 1; j <= n; j++)
		{
			if (vis[j] != 1 && t > dis[j])
			{
				t = dis[j];
				a = j;
			}
		}
		vis[a] = 1;
		for (int j = 1; j <= n; j++)
		{
			if (vis[j] != 1 && dis[j] > map[a][j] && map[a][j] != inf)
				dis[j] = map[a][j];
		}
	}
	int res = 0;
	for (int i = 1; i <= n; i++)
	{
	/*	cout << dis[i];*/
		res += dis[i];
	}
	return res;
}
int main()
{
	int v;
	cin >> v;
	n = v;
	memset(map, 0x3f, sizeof(map));
	for (int i = 1; i <= v; i++)
	{
		for (int j = 1; j <= v; j++)
		{
			cin >> map[i][j];
		}
	}
	cin >> m;
	while (m--)
	{
		int f1, f2;
		cin >> f1 >> f2;
		map[f1][f2] = map[f2][f1] = 0;
	}
	//prim();
	cout << prim();
	return 0;
}