spfa的最小生成树~

B - 基础

Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit  Status

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

这个可以利用spfa计算最小生成树 模板....

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<queue>
#include<algorithm>
using namespace std;

#define N 210
#define d double
#define INF 0xffffff
int n;
int visit[N];
d dis[N],len[N][N];
d a,b,c;

struct node
{
	d x,y;
}s[N];

double wal(int i,int j)
{
    d length=sqrt(((d)s[i].x-(d)s[j].x)*((d)s[i].x-(d)s[j].x)+((d)s[i].y-(d)s[j].y)*((d)s[i].y-(d)s[j].y));
    return 60*length/10000.0 ;
}
double sub(int i,int j)
{
    d length=sqrt(((d)s[i].x-(d)s[j].x)*((d)s[i].x-(d)s[j].x)+((d)s[i].y-(d)s[j].y)*((d)s[i].y-(d)s[j].y));
    return 60*length/40000.0;
}

void clean()
{
	for(int i=0;i<N;i++)
		for(int j=0;j<N;j++)
		len[i][j]=-1;
}

void SPFA()
{
	memset(visit,0,sizeof(visit));
	for(int i=0;i<=n;i++)
		dis[i]=INF;
	visit[1]=1;
	dis[1]=0;
	queue<int>que;
	que.push(1);
	while(!que.empty())
	{
		int t=que.front();
		for(int i=1;i<=n;i++)
		{
			if(len[t][i]!=-1 && dis[i]>dis[t]+len[t][i])
			{
				dis[i]=dis[t]+len[t][i];
				if(visit[i]==0)
				{
					visit[i]=1;
					que.push(i);
				}
			}
		}
		que.pop();
		visit[t]=0;
	}
}

int main()
{
	scanf("%lf %lf %lf %lf",&s[1].x,&s[1].y,&s[2].x,&s[2].y);
	len[1][2]=len[2][1]=wal(1,2);
	n=3;c=0;
	while(~scanf("%lf %lf",&a,&b))
	{
		if(a==-1 && b==-1)
		{
			c=0;
		}
		else
		{
			s[n].x=a;
		    s[n].y=b;
		    for(int i=1;i<n-c;i++)
				len[i][n]=len[n][i]=wal(i,n);
		    for(int i=n-c;i<n;i++)
			    len[i][n]=len[n][i]=sub(i,n);
		    c=1;
		    n++;
		}
	}
	n--;
	SPFA();
	printf("%.0lf\n",dis[2]);
    return 0;
}