poj 1258

最小生成树prim

#include <iostream>

using namespace std;

const int INF = 1000005;
const int MAXN = 105;

int cost[MAXN][MAXN];
int mincost[MAXN];
bool used[MAXN];
int n;

int prim() {
	for(int i=0; i<n; i++) {
		mincost[i] = INF;
		used[i] = false;
	}
	mincost[0] = 0;
	int res = 0;
	
	while(true) {
		int v = -1;
		for(int u=0; u<n; u++)
			if(!used[u] && (v==-1 || mincost[u]<mincost[v]))
				v = u;
		if(v == -1)	break;
		used[v] = true;
		res += mincost[v];
		for(int u=0; u<n; u++)
			mincost[u] = min(mincost[u], cost[v][u]);
	}
	
	return res;
}

int main() {
	
	while(~scanf("%d", &n)) {
		for(int i=0; i<n; i++)
		for(int j=0; j<n; j++)
			scanf("%d", &cost[i][j]);
		printf("%d\n", prim());
	}	
	
	return 0;
}


posted @ 2016-09-28 14:37  StevenLuke  阅读(137)  评论(0编辑  收藏  举报