zoukankan      html  css  js  c++  java
  • POJ 2253 Frogger(最小最大距离)

    题意  给你n个点的坐标  求第1个点到第2个点的全部路径中两点间最大距离的最小值  

    非常水的floyd咯

    #include<cstdio>
    #include<cmath>
    #include<cstring>
    #include<algorithm>
    using namespace std;
    const int N=205;
    double d[N][N];
    int x[N],y[N],n;
    
    void floyd()
    {
        for(int k=1;k<=n;++k)
        for(int i=1;i<=n;++i)
        for(int j=1;j<=n;++j)
            d[i][j]=min(d[i][j],max(d[i][k],d[k][j]));
    }
    
    int main()
    {
        int cas=0;
        while(scanf("%d",&n),n)
        {
            memset(d,0x3f,sizeof(d));
            for(int i=1;i<=n;++i)
            {
                scanf("%d%d",&x[i],&y[i]);
                for(int j=1;j<i;++j)
                {
                    int tx=x[i]-x[j],ty=y[i]-y[j];
                    d[i][j]=d[j][i]=sqrt(tx*tx+ty*ty);
                }
            }
            floyd();
            printf("Scenario #%d
    Frog Distance = %.3f
    
    ",++cas,d[1][2]);
        }
        return 0;
    }
    

    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
    

    Source



  • 相关阅读:
    单例模式
    C++中迭代器原理、失效和简单实现
    C++中静态成员变量要在类外部再定义或初始化的原因
    idea maven javaweb项目迁移时的maven和版本报错问题解决(可解决同类错误)
    java 继承类之后,访问不到超类的属性的原因及解决方法
    spring boot admin
    javaweb 报表生成(pdf excel)所需要用到的技术和思路
    团队合作开发git冲突解决方案 Intellij IDEA
    【项目管理】 使用IntelliJ IDEA 将项目发布(提交)到GitLab
    IDEA/Git 设置多个push远程仓库或者同时提交多个push仓库
  • 原文地址:https://www.cnblogs.com/slgkaifa/p/6826849.html
Copyright © 2011-2022 走看看