Constructing Roads HDU杭电1102【Kruscal || Prim】

Problem 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

//输入方式：第一行有N个村庄，然后是N行，每行是一个村庄与其他村庄的距离，然后是一个T，表示已经修好的路的条数，求最少还要修多长的路

//Kruscal

#include<cstdio>
#include<algorithm>
using namespace std;

struct node
{
	int u,v,w;
	int flag;
};

bool cmp(node a,node b)
{
	if(a.flag==b.flag) return a.w<b.w;
	return a.flag>b.flag;
}
node arr[10000];
int per[110];
int n;
void init()
{
	for(int i=1;i<=n;++i)
	{
		per[i]=i;
	}
}

int find(int x)
{
	if(x==per[x]) return x;
	return per[x]=find(per[x]);
}
bool join(int x,int y,int f)
{
	int fx=find(x);
	int fy=find(y);
	if(fx!=fy)
	{
		per[fx]=fy;
		if(f==1) return 0;
		
		return 1;
	}
	return 0;
}
int main()
{
	int a,b,temp;
	int T;
	int i,j;
	while(~scanf("%d",&n))
	{
		init();
		int k=0;
		for(i=1;i<=n;++i)
		{
			for(j=1;j<=n;++j)
			{
				scanf("%d",&temp);
				if(temp)
				{
					arr[k].u=i;
					arr[k].v=j;
					arr[k].flag=0;
					arr[k++].w=temp;
				}
			}
		}
		scanf("%d",&T);
		while(T--)
		{
			scanf("%d%d",&a,&b);
			for(i=0;i<k;++i)
			{
				if(arr[i].u==a&&arr[i].v==b)
				{
					arr[i].flag=1;
					break;
				}
			}
		}
		sort(arr,arr+k,cmp);
		int sum=0;
		for(i=0;i<k;++i)
		{
			if(join(arr[i].u,arr[i].v,arr[i].flag))
			{
				sum+=arr[i].w;
			}
		}
		printf("%d\n",sum);
	}
	return 0;
}

//prim

#include<stdio.h>
#include<string.h>
#define INF 0x3f3f3f3f
#define N 110
int n;
int i,j;
int map

;
int a,b;
int low
;
bool vis
;
void input()
{
	for(i=1;i<=n;++i)
	{
		for(j=1;j<=n;++j)
		{
			scanf("%d",map[i]+j);
		}
	}
	int m;
	scanf("%d",&m);
	while(m--)
	{
		scanf("%d%d",&a,&b);
		//vis[a]=vis[b]=1;//把已经连接了的两个点加入最小生成树集合 
		map[a][b]=map[b][a]=0; //把ab两点的权值变为0 
	}
}
void prim()
{
	int pos=1;
	int sum=0; 
	for(i=1;i<=n;++i)//第一次给low赋值 
	{
		low[i]=map[pos][i];
	}
	vis[pos]=1; //加入最小生成树集合 
	for(i=1;i<n;++i)//再找n-1个点 
	{
		int min=INF;
		for(j=1;j<=n;++j)
		{
			if(!vis[j]&&min>low[j])
			{
				min=low[j];
				pos=j;//把找到的点记录下
			}
		}
		sum+=min;
		vis[pos]=1; 
		for(j=1;j<=n;++j)
		{
			if(!vis[j]&&low[j]>map[pos][j])
			{
				low[j]=map[pos][j];
			}
		} 
	}
	printf("%d\n",sum);
}
int main()
{
	
	while(~scanf("%d",&n))
	{
		memset(vis,0,sizeof(vis));
		input();
		prim();
	}
	return 0;
}