POJ-2421-Constructing Roads（最小生成树 普利姆）

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 26694		Accepted: 11720

 

Description

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

题意：题意十分简单粗暴，首行给出N，接着是个N*N的矩阵，map[i][j]就代表i到j的权值。接着给出Q，下面Q行，每行两个数字A，B，代表A到B，B到A的权值为0。最后输出最小生成树的权值和就行。

思路：由于是稠密图，所以选用普利姆算法比较合适，

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
#define N 110
#define M 0x3f3f3f3f//用一个大值表示两点不通 

int map[N][N];
int vis[N],dst[N];//vis标已加入MST的点，dst存放各点到MST的最小距离 
int n,q;

void init()//初始图 
{
	int i,j;
	
	for (i=0;i<N;i++)
	{
		for (j=0;j<N;j++)
		{
			i==j?map[i][j]=0:map[i][j]=M;//自己到自己的点距离为0 
		}
	}
	memset(vis,0,sizeof(vis));
	memset(dst,0,sizeof(dst));
}

void prime()
{
	int ans=0,i,min,j,k,point;
	
	vis[1]=1;//1放入MST 
	for (i=1;i<=n;i++)
	{
		dst[i]=map[i][1];//dst初始化 
	}
	
	for (i=1;i<=n;i++)
	{
		min=M;
		for (j=1;j<=n;j++)//找距MST最近的点 
		{
			if (vis[j]==0&&min>dst[j])
			{
				min=dst[j];
				point=j;
			}
		}
		if (min==M)//没有连通点 
		{
			break;
		 }
		 
		vis[point]=1;//把距MST最近的点加入MST 
		ans=ans+dst[point];
		
		for (k=1;k<=n;k++)//更新各点到MST的最小距离 
		{
			if (vis[k]==0&&dst[k]>map[k][point])
			{
				dst[k]=map[k][point];
			} 
		}
	}
	printf("%d
",ans);
}

int main()
{
	int i,j,x,y;
	
	scanf("%d",&n);
	init();
	for (i=1;i<=n;i++)
	{
		for (j=1;j<=n;j++)
		{
			scanf("%d",&map[i][j]);
		}
	}
	
	scanf("%d",&q);
	for (i=0;i<q;i++)
	{
		scanf("%d%d",&x,&y);//已连通的两点权为0 
		map[x][y]=0;
		map[y][x]=0;
	}
	
	prime();
	
	return 0;
}