POJ 2253 Frogger(最短路径Dijkstra的变形)

Frogger

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 15491		Accepted: 5124

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

 解题报告：给出两只青蛙的起始位置，找出青蛙跳到第2块的最短距离。这个题就是最短路径的变形，就是在求最短路径时更新最短路信息的时候不一样，

即在每条路径里最短的最长的那条分路径，再在里面选择最短的那条。

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
const int N = 210;
const double INF = 99999999;
double map[N][N], dis[N];
int visit[N], n;
struct node
{
    double x, y;
}a[N];
double Dis(int i, int j)//计算两点间的距离
{
    return sqrt((a[i].x - a[j].x) * (a[i].x - a[j].x) + (a[i].y - a[j].y) * (a[i].y - a[j].y));
}
double Max(double c, double d)
{
    return c > d ? c : d;
}
double Min(double c, double d)
{
    return c < d ? c : d;
}
void Dijkstra()
{
    int i, j, p;
    double min, max;
    for (i = 1; i <= n; ++i)//初始化
    {
        dis[i] = map[1][i];
    }
    visit[1] = 1;
    for (i = 2; i <= n; ++i)
    {
        min = INF;
        for (j = 1; j <= n; ++j)
        {
            if (!visit[j] && min > dis[j])
            {
                min = dis[j];
                p = j;
            }
        }
        visit[p] = 1;
        for (j = 1; j <= n; ++j)//就是这儿和Dijkstra算法不同，
        {
            max = Max(dis[p], map[p][j]);//每条路径里最短的最长的那条分路径
            dis[j] = Min(dis[j], max);
        }
    }
}
int main()
{
    int i, j, count = 1;
    while (scanf("%d", &n) != EOF && n)
    {
        memset(map, 0, sizeof(map));
        memset(visit, 0, sizeof(visit));
        memset(dis, 0, sizeof(dis));
        for(i = 1; i <= n; ++i)
        {
            scanf("%lf%lf", &a[i].x, &a[i].y);
        }
        for (i = 1; i <= n; ++i)
        {
            for (j = 1; j <= i - 1; ++j)
            {
                map[i][j] = map[j][i] = Dis(i, j);
            }
        }
        Dijkstra();
        printf("Scenario #%d\n", count++);
        printf("Frog Distance = %.3lf\n\n", dis[2]);
    }
}