还是畅通工程

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 25591    Accepted Submission(s): 11370

3
1 2 1
1 3 2
2 3 4
4
1 2 1
1 3 4
1 4 1
2 3 3
2 4 2
3 4 5
0

3
5

Hint
Hint
 
Huge input, scanf is recommended.

第一次写最小生成树，代码很那啥天然无雕饰哈（很费时..）

#include <stdio.h>
#include <string.h>
#define maxn 102

int map[maxn][maxn];
bool vis[maxn];

int Prim(int n)
{
	int len = 0, i, j, tmp, u, v, count = 0;
	vis[1] = true;
	while(count < n - 1){
		for(i = 1, tmp = -1; i <= n; ++i){
			for(j = 1; vis[i] && j <= n; ++j) //cut
				if(map[i][j] != -1 && !vis[j] && (tmp == -1 || map[i][j] < tmp)){
					tmp = map[i][j]; u = j; v = i;
				}			
		}
		if(tmp != -1){
			map[v][u] = -1;
			len += tmp; ++count;
			vis[u] = 1;
		}
	}
	return len;
}

int main()
{
	//freopen("in.txt", "r", stdin);
	//freopen("out.txt", "w", stdout);
	int n, m, a, b, c, i;
	while(scanf("%d", &n), n){
		memset(map, -1, sizeof(map));
		memset(vis, 0, sizeof(vis));
		m = n * (n - 1) / 2;
		for(i = 0; i < m; ++i){
			scanf("%d%d%d", &a, &b, &c);
			if(map[a][b] == -1 || c < map[a][b])
				map[a][b] = map[b][a] = c;
		}
		printf("%d\n", Prim(n));
	}
	return 0;
}

HDU1233 还是畅通工程 【最小生成树Prim】,布布扣,bubuko.com

HDU1233 还是畅通工程 【最小生成树Prim】

原文：http://blog.csdn.net/chang_mu/article/details/38256587