Frogger

Frogger

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 38575		Accepted: 12424

Description

Freddy Frog is sitting on a stone in the middle of a lake. Suddenly he notices Fiona Frog who is sitting on another stone. He plans to visit her, but since the water is dirty and full of tourists' sunscreen, he wants to avoid swimming and instead reach her by jumping.
Unfortunately Fiona's stone is out of his jump range. Therefore Freddy considers to use other stones as intermediate stops and reach her by a sequence of several small jumps.
To execute a given sequence of jumps, a frog's jump range obviously must be at least as long as the longest jump occuring in the sequence.
The frog distance (humans also call it minimax distance) between two stones therefore is defined as the minimum necessary jump range over all possible paths between the two stones.

You are given the coordinates of Freddy's stone, Fiona's stone and all other stones in the lake. Your job is to compute the frog distance between Freddy's and Fiona's stone.

Input

The input will contain one or more test cases. The first line of each test case will contain the number of stones n (2<=n<=200). The next n lines each contain two integers xi,yi (0 <= xi,yi <= 1000) representing the coordinates of stone #i. Stone #1 is Freddy's stone, stone #2 is Fiona's stone, the other n-2 stones are unoccupied. There's a blank line following each test case. Input is terminated by a value of zero (0) for n.

Output

For each test case, print a line saying "Scenario #x" and a line saying "Frog Distance = y" where x is replaced by the test case number (they are numbered from 1) and y is replaced by the appropriate real number, printed to three decimals. Put a blank line after each test case, even after the last one.

Sample Input

2
0 0
3 4

3
17 4
19 4
18 5

0

Sample Output

Scenario #1
Frog Distance = 5.000

Scenario #2
Frog Distance = 1.414

Source

Ulm Local 1997

/*
给出你n个点的坐标（二维平面中的），让你求这样一个值：从点1到点2有很多路径，中间可以从很多块石头跳过去，现在就让你找一条路径，从点1
到点2，中间每跳一步中的最大距离d，是所有路径中最小的。

跟Kruskal算法差不多，先按照边的长短来排序，然后每次增加一条边，判断是不是连通，如果连通就是找到的最短路；
*/
#include <iostream>
#include <stdio.h>
#include <math.h>
#include <string.h>
#include <string>
#include <algorithm>
#define N 205
using namespace std;
struct node
{
    int x,y;
    double d;//两个点之间的距离
    bool operator <(const node &b) const
    {
        return d<b.d;
    }//按照边的长短排序
};
node fr[N*N];
int bin[N];
int nod[N];
int m=0;
int n;
int findx(int x)
{
    int child=x;
    while(x!=bin[x])
        x=bin[x];
    while(child!=x)
    {
        int t=bin[child];
        bin[child]=x;
        child=t;
    }
    return x;
}
void built(int x,int y)
{
    int fx=findx(x);
    int fy=findx(y);
    if(fx!=fy)
        bin[fx]=fy;
}
double dis(int x1,int y1,int x2,int y2)
{
    return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
}
void init()
{
    for(int i=1;i<=n;i++)
        bin[i]=i;
}
int main()
{
    freopen("C:\\Users\\acer\\Desktop\\in.txt","r",stdin);
    int Case=0;
    while(scanf("%d",&n)!=EOF&&n)
    {
        if(Case)
            printf("\n");
        printf("Scenario #%d\n",++Case);
        m=0;
        int x[N],y[N];
        for(int i=1;i<=n;i++)
            scanf("%d%d",&x[i],&y[i]);
        for(int i=1;i<=n;i++)
        {
            for(int j=i+1;j<=n;j++)
            {
                fr[m].x=i;
                fr[m].y=j;
                fr[m++].d=dis(x[i],y[i],x[j],y[j]);
            }
        }
        sort(fr,fr+m);
        
        for(int i=0;i<m;i++)
        {
            int flag=0;
            init();
            double answer=fr[i].d;
            for(int j=i;j<m;j++)
            {
                answer=fr[j].d;
                built(fr[j].x,fr[j].y);
                if(findx(1)==findx(2))
                {
                    printf("Frog Distance = %.3f\n",answer);
                    flag=1;
                    break;
                }
            }
            if(flag)
                break;
        }
    }
    return 0;
}

 

转载于:https://www.cnblogs.com/wuwangchuxin0924/p/5942462.html