最短路练习2/poj/2253/Frogger

题目链接：http://poj.org/problem?id=2253

Frogger

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 42941 Accepted: 13693

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

20 03 4317 419 418 50

Sample Output

Scenario #1Frog Distance = 5.000Scenario #2Frog Distance = 1.414

题意：给出n个点的坐标，求点2到点1跳的最小必要跳跃距离（低于这个距离就不可能跳的点1）。

AC代码：

#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<algorithm>#include<cmath>#include<queue>#include<vector>#include<map>#include<string>#define LL long longusing namespace std;const int mod = 1e7+7;const int inf = 0x3f3f3f3f;const int maxn = 1e6 +10;double ma[205][205];//存两点间距离int x[205],y[205];//坐标int vis[205];//标记是否更新为最小值double dis[205];//dis[i]表示i点到1点的最小必要跳跃距离。double sqr;int n;int main(){    int flag=0;    while(~scanf("%d",&n)&&n)    {        memset(vis,0,sizeof(vis));         for(int i=1;i<=n;i++)         {             scanf("%d%d",&x[i],&y[i]);             for(int j=1;j<=i;j++)             {                 sqr=sqrt((double)(x[i]-x[j])*(double)(x[i]-x[j])+(double)(y[i]-y[j])*(double)(y[i]-y[j]));                 ma[i][j]=sqr;                 ma[j][i]=sqr;             }         }         for(int i=1;i<=n;i++)         {             dis[i]=ma[i][1];         }         int nn=n-2;         vis[1]=1;         while(nn--)         {             double sum=inf;             int point=-1;             for(int i=2;i<=n;i++)             {                 if(!vis[i]&&dis[i]<sum)                 {                     sum=dis[i];                     point=i;                 }             }             if(point!=-1)             {                 vis[point]=1;                 for(int i=2;i<=n;i++)                 {                     if(!vis[i]&&max(dis[point],ma[point][i])<dis[i])//画个三角形表示三条边长，你就知道为什么这样更新了。                        dis[i]=max(dis[point],ma[point][i]);                 }             }         }         if(flag++)            printf("\n");         printf("Scenario #%d\nFrog Distance = %.3lf\n",flag,dis[2]);    }}