poj-2253-Frogger

ゝ一世哀愁。 2022-05-28 12:07 113阅读 0赞

Frogger

<table> 
 <tbody> 
  <tr> 
   <td><strong>Time Limit:</strong>&nbsp;1000MS</td> 
   <td>&nbsp;</td> 
   <td><strong>Memory Limit:</strong>&nbsp;65536K</td> 
  </tr> 
  <tr> 
   <td><strong>Total Submissions:</strong>54307</td> 
   <td>&nbsp;</td> 
   <td><strong>Accepted:</strong>&nbsp;17140</td> 
  </tr> 
 </tbody> 
</table>

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：给出青蛙的坐标，以及各个石块的坐标，求1号青蛙到2号青蛙之间至少需要跳的最大距离，不是最短路问题，路可以很长，跳的石头很多，要求是跳的最大距离，最小；

代码：

#include<queue>
    #include<math.h>
    #include<stdio.h>
    #include<string.h>
    #include<stdlib.h>
    #include<algorithm>
    using namespace std;
    #define inf 99999999
    double e[1010][1010];
    int book[1010];
    double dis[1010];
    struct point
    {
        int x,y;
        double z;
    } a[10000];
    double di(point A,point B)
    {
        return sqrt((double)((A.x-B.x)*(A.x-B.x)+(A.y-B.y)*(A.y-B.y)));
    }
    int main()
    {
        int n,q=1;
        while(scanf("%d",&n)!=EOF)
        {
            int i,j,k,u,v,w;
            double minn;
            if(n==0)
                break;
            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",&a[i].x,&a[i].y);
            }
            for(i=1; i<=n; i++)
            {
                for(j=i+1; j<=n; j++)
                {
                    double d=di(a[i],a[j]);
                        e[i][j]=e[j][i]=d;
                }
            }
            for(i=1; i<=n; i++)
                dis[i]=e[1][i];
            for(i=1; i<=n; i++)
                book[i]=0;
            book[1]=1;
            for(i=1; i<=n-1; i++)
            {
                minn=inf;
                for(j=1; j<=n; j++)
                {
                    if(book[j]==0&&dis[j]<minn)
                    {
                        minn=dis[j];
                        u=j;
    
                    }
                }
                book[u]=1;
                for(v=1; v<=n; v++)
                {
                    dis[v]=min(dis[v],max(dis[u],e[u][v]));
                }
            }
            printf("Scenario #%d\n",q++);
            printf("Frog Distance = %.3lf\n\n",dis[2]);
        }
        return 0;
    }

[Ulm Local 1997]: http://poj.org/searchproblem?field=source&key=Ulm+Local+1997