POJ2253——Frogger(Floyd变形)

Frogger

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

题目大意：

　　　　一只小青蛙想要去见另外一只小青蛙。他在一块石头上，她在另外一块石头上。由于河水污染严重，他每次在河水中待的时间不能太长。

　　　　有N块石头，他在N1上，她在N2上，选择最合适的路线，使他每次在河水中停留的时间最短。

解题思路：

　　　　Floyd变形。

核心代码：

if (edge[i][j]>edge[i][m]&&edge[i][j]>edge[m][j])
    edge[i][j]=max(edge[i][m],edge[m][j]);

Edge里面存的当前路径中最大的停留时间。

Code：

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<math.h>
#define MAXN 1000
using namespace std;
struct point
{
    double x,y;
} P[MAXN+10];
double edge[MAXN+10][MAXN+10];
double dis2p(struct point a,struct point b)
{
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
double max(double a,double b)
{
    return a>b?a:b;
}
int main()
{
    int N,times=0;
    while (cin>>N)
    {
        times++;
        if (N==0) break;
        for (int i=1; i<=N; i++)
            cin>>P[i].x>>P[i].y;
        for (int i=1; i<=N; i++)
            for (int j=1; j<=N; j++)
                edge[i][j]=dis2p(P[i],P[j]);
        for (int m=1; m<=N; m++)
            for (int i=1; i<=N; i++)
                for (int j=1; j<=N; j++)
                {
                    //if (j==i) continue;
                    //if (m==j||m==i) continue;
                    if (edge[i][j]>edge[i][m]&&edge[i][j]>edge[m][j])
                        edge[i][j]=max(edge[i][m],edge[m][j]);
                }
        printf("Scenario #%d
",times);
        printf("Frog Distance = %.3lf

",edge[1][2]);
    }
    return 0;
}