51Nod: 1212 无向图最小生成树。
link： http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1212
第1行：2个数N,M中间用空格分隔，N为点的数量，M为边的数量。（2 <= N <= 1000, 1 <= M <= 50000) 第2 - M + 1行：每行3个数S E W，分别表示M条边的2个顶点及权值。(1 <= S, E <= N，1 <= W <= 10000)
输出最小生成树的所有边的权值之和。
9 14 1 2 4 2 3 8 3 4 7 4 5 9 5 6 10 6 7 2 7 8 1 8 9 7 2 8 11 3 9 2 7 9 6 3 6 4 4 6 14 1 8 8
37
题解：
无向图最小生成树， 采用prime算法。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
const int INF = 0x3f3f3f3f;
const int maxn = 1005;
int n,m,mp[maxn][maxn], dist[maxn], vis[maxn];
int prime(int start){
memset(vis, 0, sizeof(vis));
memset(dist, INF, sizeof(dist));
vis[start] = 1;
int pt, mindist, path = 0, cur = start;
for(int i=2; i<=n; ++i){
mindist = INF;
pt = -1;
for(int i=1; i<=n; ++i){
if(!vis[i] && mp[cur][i] > 0 && mp[cur][i] < dist[i]){
dist[i] = mp[cur][i];
}
if(!vis[i] && mindist > dist[i]){
mindist = dist[i];
pt = i;
}
}
path += mindist;
vis[pt] = 1;
cur = pt;
}
return path;
}
int main(){
freopen("in.txt", "r", stdin);
int x, y, val, ans;
while(scanf("%d %d", &n, &m) != EOF){
memset(mp, 0, sizeof(mp));
for(int i=0; i<m; ++i){
scanf("%d %d %d", &x, &y, &val);
mp[x][y] = mp[y][x] = val;
}
ans = prime(1);
printf("%d\n", ans );
}
return 0;
}
原文：http://www.cnblogs.com/zhang-yd/p/6011247.html