zoukankan      html  css  js  c++  java
  • hdu 4709 Herding

    Herding

    Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 886 Accepted Submission(s): 240

    Problem Description
    Little John is herding his father's cattles. As a lazy boy, he cannot tolerate chasing the cattles all the time to avoid unnecessary omission. Luckily, he notice that there were N trees in the meadow numbered from 1 to N, and calculated their cartesian coordinates (Xi, Yi). To herding his cattles safely, the easiest way is to connect some of the trees (with different numbers, of course) with fences, and the close region they formed would be herding area. Little John wants the area of this region to be as small as possible, and it could not be zero, of course.
     
    Input
    The first line contains the number of test cases T( T<=25 ). Following lines are the scenarios of each test case.
    The first line of each test case contains one integer N( 1<=N<=100 ). The following N lines describe the coordinates of the trees. Each of these lines will contain two float numbers Xi and Yi( -1000<=Xi, Yi<=1000 ) representing the coordinates of the corresponding tree. The coordinates of the trees will not coincide with each other.
     
    Output
    For each test case, please output one number rounded to 2 digits after the decimal point representing the area of the smallest region. Or output "Impossible"(without quotations), if it do not exists such a region.
     
    Sample Input
    1 4 -1.00 0.00 0.00 -3.00 2.00 0.00 2.00 2.00
     
    Sample Output
    2.00
     
    Source
     
    Recommend
    liuyiding
    就是求三角形的最小面积,不能用海伦公式,估计是有精度损失吧!
    #include <iostream>
    #include <math.h>
    #include <stdio.h>
    #include <string.h>
    using namespace std;
    #define inf 100000000.0
    double x[105],y[105];
    double ff(int i,int j,int k)
    {
        return fabs(x[i]*y[j]+y[i]*x[k]+x[j]*y[k]-x[k]*y[j]-x[j]*y[i]-x[i]*y[k])/2;
    }
    int main()
    {
        int tcase,i,n,j,k;
        scanf("%d",&tcase);
        while(tcase--)
        {
            scanf("%d",&n);
            for(i=0;i<n;i++)
            {
                scanf("%lf%lf",&x[i],&y[i]);
            }
            double maxx=inf;
            bool falg=true;
            for(i=0;i<n-2;i++)
                for(j=i+1;j<n-1;j++)
                    for(k=j+1;k<n;k++)
                    {
                        double tt=ff(i,j,k);
                        if(tt!=0.0)
                        maxx=min(maxx,tt),falg=false;
                    }
            if(!falg)
            printf("%.2lf
    ",maxx);
            else
            printf("Impossible
    ");
        }
        return 0;
    }
    


     
  • 相关阅读:
    【OpenStack】OpenStack系列16之OpenStack镜像制作
    【OpenStack】OpenStack系列15之OpenStack高可用详解
    【OpenStack】OpenStack系列14之Dashboard定制开发
    【OpenStack】OpenStack系列13之Nova源码解析与API扩展
    linux之scp命令
    阿里大鱼短信平台
    count(1)与count(*)
    java中枚举类到高级使用
    idea中mybatis-plugin破解
    mmall项目之问题一(mavenplugin问题)
  • 原文地址:https://www.cnblogs.com/riskyer/p/3313314.html
Copyright © 2011-2022 走看看