POJ 2253 Frogger（dijkstra最短路）_poj2253 frogger dijkstra-优快云博客

Frogger

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

Sample Output

Scenario #1
Frog Distance = 5.000

Scenario #2
Frog Distance = 1.414

题意描述：
给出各个石头的坐标，其中第一块石头为青蛙起点，第二块石头为青蛙终点。现青蛙可以借助其他石头到达第二块石头，问青蛙的跳远极限距离最小为多少。（跳跃极限距离为青蛙在路径中跳跃的最大距离）

解题思路：
将dijkstra最短路算法稍微改变，将dis数组用来存路径中跳远极限距离的最小值。
注意每个例子之间有两个换行。

AC代码：

#include<stdio.h>
#include<string.h>
#include<math.h>
int book[1010],inf=99999999,min,x[1010],y[1010];
double e[1010][1010],dis[1010];
int main(void)
{
	int i,j,k,n,m,t=0,u;
	while(scanf("%d",&n)!=EOF)
	{
		if(n==0)
		break;
		memset(book,0,sizeof(book));
		memset(dis,0,sizeof(dis));
		memset(x,0,sizeof(x));
		memset(y,0,sizeof(0));
		for(i=1;i<=n;i++)
		for(j=1;j<=n;j++)
		{
			if(i==j)
			e[i][j]=0;
			else
			e[i][j]=inf;
		}
		for(i=1;i<=n;i++)
		{
			scanf("%d%d",&x[i],&y[i]);
		}
		for(i=1;i<=n;i++)
		for(j=1;j<=n;j++)
		{
			e[i][j]=sqrt(1.0*(x[i]-x[j])*(x[i]-x[j])+1.0*(y[i]-y[j])*(y[i]-y[j])); 
		}
		for(i=1;i<=n;i++)
		dis[i]=e[1][i];
		book[1]=0;
		for(i=1;i<=n;i++)
		{
			min=inf;
			for(j=1;j<=n;j++)
			{
				if(dis[j]<min&&book[j]==0)
				{
					u=j;
					min=dis[j];
				}
			}
			book[u]=1;
			for(k=1;k<=n;k++)
			{
				if(e[u][k]<inf)
				{
					if(dis[k]>dis[u]&&dis[u]>=e[u][k])
					dis[k]=dis[u];
					else if(dis[k]>e[u][k]&&e[u][k]>=dis[u])
					dis[k]=e[u][k];
				}
			}
		}
		printf("Scenario #%d\n",++t);
		printf("Frog Distance = %.3lf\n\n",dis[2]); 
	}
 }