POJ 2253 Frogger

Frogger

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 19928		Accepted: 6464

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

 

 

题目大意:

给出两只青蛙的坐标A、B，和其他的n-2个坐标，任一两个坐标点间都是双向连通的。显然从A到B存在至少一条的通路，每一条通路的元素都是这条通路中前后两个点的距离，这些距离中又有一个最大距离。

现在要求求出所有通路的最大距离，并把这些最大距离作比较，把最小的一个最大距离作为青蛙的最小跳远距离。

 

Floyd算法

用Floyd算法求出两两最短路，再求出从每个点开始的最长路，最后从这n个最长路中求出最小的那个即为所求。

 

 Floyd:

 

#include<iostream>

#include<cstdio>

#include<cstring>

#include<cmath>



using namespace std;



struct node{

    int x,y;

}point[210];



double dis[210][210];



double Cal(node a,node b){

    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));

}



int main(){



    //freopen("input.txt","r",stdin);



    int n,cases=0;

    while(~scanf("%d",&n) && n){

        for(int i=1;i<=n;i++)

            scanf("%d%d",&point[i].x,&point[i].y);

        for(int i=1;i<=n;i++)

            for(int j=i+1;j<=n;j++)

                dis[i][j]=dis[j][i]=Cal(point[i],point[j]);

        for(int k=1;k<=n;k++)

            for(int i=1;i<=n;i++)    //主要针对由i到j的松弛,最终任意两点间的权值都会被分别松弛为最大跳的最小（但每个两点的最小不一定相同）

                for(int j=1;j<=n;j++)

                    if(dis[i][k]<dis[i][j] && dis[k][j]<dis[i][j])   //当边ik,kj的权值都小于ij时，则走i->k->j路线，否则走i->j路线   

                        dis[i][j]=dis[j][i]=max(dis[i][k],dis[k][j]);    //当走i->k->j路线时，选择max{ik,kj},只有选择最大跳才能保证连通

        printf("Scenario #%d\n",++cases);

        printf("Frog Distance = %.3f\n\n",dis[1][2]);

    }

    return 0;

}

 

 

#include <stdio.h>

#include <math.h>

#define MAX 1000000

#define M 205

struct node

{

    int x,y;

}stone[M];



double edge[M][M];

double dis(node a,node b)     //算两点间的距离

{

    double len = (a.x-b.x)*(a.x-b.x) + (a.y-b.y)*(a.y-b.y);

    len = sqrt (len);

    return len;

}



double prim (int n)

{

    int start = 0,end = 1;     //起始和终了位置

    int mark[M],k,i,j;

    double d[M],ans;

    for (i = 0;i < n;i ++)

    {

        mark[i] = 0;

        d[i] = edge[start][i];

    }

    mark[start] = 1;

    ans = 0;

    for (i = 1;i < n;i ++)

    {

        double min = MAX;

        for (j = 1;j < n;j ++)

            if (!mark[j]&&d[j] < min)

            {

                min = d[j];

                k = j;

            }

        mark[k] = 1;

        ans = ans > d[k]?ans:d[k];     //最知道路径中最大的

        if (k == end) break;           //到终点了退出

        for (j = 1;j < n;j ++)

            if (d[j] > edge[k][j])

                d[j] = edge[k][j];

    }

    return ans;

}

int main ()

{

    int n,count = 1;

    int i,j,k;

    while (scanf ("%d",&n)&&n)

    {

        for (i = 0;i < n;i ++)

            for (j = 0;j < n;j ++)

                edge[i][j] = MAX;

        for (i = 0;i < n;i ++)

            scanf ("%d%d",&stone[i].x,&stone[i].y);

        k = 0;

        for (i = 0;i < n;i ++)

            for (j = i+1;j < n;j ++)

            {

                double len = dis (stone[i],stone[j]);

                edge[i][j] = len;

                edge[j][i] = len;

            }

        double ans = prim(n);

        printf ("Scenario #%d\n",count++);

        printf ("Frog Distance = %.3f\n\n",ans); //POJ上不能用%.3lf,,这题WA到我崩溃。。。。。

    }

}