HDOJ 1863:畅通工程 Prim算法求解最小生成树

    http://acm.hdu.edu.cn/showproblem.php?pid=1863

    这道浙大的考研上机题题意大致是：给一个带全图，若连图，则输出该图的最小生成树的权值之和，不连通，输出？.

    我用Prim算法求解MST，求解然之后判断是否连通，判断连通并没有去重新遍历图一遍。用了一个小技巧。

    我的AC代码。

   

#include<iostream>
#include<stdio.h>
using namespace std;

const int Infinity = 1000000000;
const int Max = 100 + 10;
int g[Max][Max];
int edges, vers;

struct Node
{
	int adj;
	int closeDist;
	bool along;
};

Node Mst[Max];

void Prim()
{
	Mst[1].along = true;
	
	for(int i=2; i<=vers; i++)
	{
		Mst[i].adj = 1;
		Mst[i].closeDist = g[1][i];
		Mst[i].along = false;
	}
	
	int min, pos;
	for(int i=1; i<vers; i++)
	{
		min = Infinity + 1, pos = 0;
		for(int j=1; j<=vers; j++)
		{
			if(!Mst[j].along && min > Mst[j].closeDist)
			{
				min = Mst[j].closeDist;
				pos = j;
			}
		}
		
		Mst[pos].along = true;
		
		for(int j=1; j<=vers; j++)
		{
			if(!Mst[j].along && g[j][pos] < Mst[j].closeDist)
			{
				Mst[j].closeDist = g[j][pos];
				Mst[j].adj = pos;
			}
		}
	}
}

int main()
{
	int head, tail, cost;
	
	while(scanf("%d%d", &edges, &vers) && edges)
	{
		for(int i=0; i<Max; i++)
			for(int j=0; j<Max; j++) g[i][j] = Infinity;
			
		for(int i=0; i<edges; i++)
		{
			scanf("%d%d%d", &head, &tail, &cost);
			g[head][tail] = g[tail][head] = cost;
		}
		
		Prim();
		
		bool connected = true;
		int sum = 0;
		
		for(int i=1; i<=vers; i++)
		{
			if(Mst[i].closeDist == Infinity)
			{
				connected = false;
				break;
			}
			else sum += Mst[i].closeDist;
		}
		
		if(connected) printf("%d\n", sum);
		else printf("?\n");
	}
	system("pause");
	return 0;
}