1405 奶牛的旅行

题目描述 Description

农民John的农场里有很多牧区。有的路径连接一些特定的牧区。一片所有连通的牧区称为一个牧场。但是就目前而言，你能看到至少有两个牧区通过任何路径都不连通。这样，农民John就有多个牧场了。 

John想在农场里添加一条路径(注意，恰好一条)。对这条路径有以下限制： 

一个牧场的直径就是牧场中最远的两个牧区的距离(本题中所提到的所有距离指的都是最短的距离)。考虑如下的有5个牧区的牧场，牧区用“*”表示，路径用直线表示。每一个牧区都有自己的坐标： 

　　　　　　 15,15 20,15
　　　　　　　　 D　　 E
　　　　　　　　 *-------*
　　　　　　　　 |　　 _/|
　　　　　　　　 | _/ |
　　　　　　　　 | _/　　|
　　　　　　　　 |/　　 |
　　　　*--------*-------*
　　　　A　　　　B　　 C
　　　　10,10 15,10 20,10
这个牧场的直径大约是12.07106, 最远的两个牧区是A和E，它们之间的最短路径是A-B-E。 

这里是另一个牧场： 

　　　　　　　　　　　　 *F 30,15
　　　　　　　　　　　　/ 
　　　　　　　　　　 _/ 
　　　　　　　　　　_/　　
　　　　　　　　 /　　 
　　　　　　　　 *------* 
　　　　　　　　 G　　 H
　　　　　　　　 25,10 30,10
这两个牧场都在John的农场上。John将会在两个牧场中各选一个牧区，然后用一条路径连起来，使得连通后这个新的更大的牧场有最小的直径。 

注意，如果两条路径中途相交，我们不认为它们是连通的。只有两条路径在同一个牧区相交，我们才认为它们是连通的。 

输入文件包括牧区、它们各自的坐标，还有一个如下的对称邻接矩阵： 

　 A B C D E F G H 
A 0 1 0 0 0 0 0 0
B 1 0 1 1 1 0 0 0
C 0 1 0 0 1 0 0 0
D 0 1 0 0 1 0 0 0
E 0 1 1 1 0 0 0 0
F 0 0 0 0 0 0 1 0
G 0 0 0 0 0 1 0 1
H 0 0 0 0 0 0 1 0
输入文件至少包括两个不连通的牧区。 

请编程找出一条连接两个不同牧场的路径，使得连上这条路径后，这个更大的新牧场有最小的直径。

输入描述 Input Description

第1行: 一个整数N (1 <= N <= 150), 表示牧区数 

第2到N+1行: 每行两个整数X,Y (0 <= X ,Y<= 100000), 表示N个牧区的坐标。注意每个 牧区的坐标都是不一样的。 

第N+2行到第2*N+1行: 每行包括N个数字(0或1) 表示如上文描述的对称邻接矩阵。

输出描述 Output Description

只有一行，包括一个实数，表示所求直径。数字保留六位小数。

样例输入 Sample Input

8
10 10
15 10
20 10
15 15
20 15
30 15
25 10
30 10
01000000
10111000
01001000
01001000
01110000
00000010
00000101
00000010

样例输出 Sample Output

22.071068

数据范围及提示 Data Size & Hint

1s

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#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<cmath>
using namespace std;
double maxn=1e12;
double x[1001];
double y[1001];
double cd(int i,int j)
{
    return sqrt((x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j]));
}
double map[1001][1001];
double m[10001];//点i可以到达的最大距离 
int main()
{
    int n;
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        scanf("%lf%lf",&x[i],&y[i]);
    }
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=n;j++)
        {
            char kk;
            //scanf("%c",&kk);
            cin>>kk;
            if(kk=='1')
            {
                double tmp=cd(i,j);
                map[i][j]=tmp;
            }
            else
            {
                map[i][j]=maxn;
            }
        }
    }
    for(int k=1;k<=n;k++)
    {
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
            {
                if(i!=j&&i!=k&&j!=k)
                {
                    if(map[i][k]<maxn-1&&map[k][j]<maxn-1)
                    {
                        if(map[i][j]>map[i][k]+map[k][j])
                        {
                            map[i][j]=map[i][k]+map[k][j];
                        }
                    }
                    
                }
                
            }
        }
    }
    memset(m,0,sizeof(m));
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=n;j++)
        {
            if(map[i][j]<maxn-1)
            {
                if(m[i]<map[i][j])
                {
                    m[i]=map[i][j];
                }
            }
            
        }
    } 
    double r2=maxn;
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=n;j++)
        {
            if(map[i][j]>maxn-1&&i!=j)
            {
                double tmp=cd(i,j);
                double nn=m[i]+m[j]+tmp;
                if(r2>nn)
                r2=nn;
            }
        }
    }
    double r1=-1;
    for(int i=1;i<=n;i++)
    {
        
        if(m[i]>r1)
        r1=m[i];
    }
    printf("%.6lf",max(r1,r2));
    return 0;
}