zoukankan      html  css  js  c++  java
  • hdu4709求三角形面积

    Herding

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 702    Accepted Submission(s): 174

    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

    分析:求最小面积就是求所有点构成的所有三角形的最小面积,但是要注意选择构成三角形的三个点不能在一条线上,横,竖,斜不能在一条线上

    #include<iostream>
    #include<cstdio>
    #include<cstdlib>
    #include<cstring>
    #include<string>
    #include<queue>
    #include<algorithm>
    #include<map>
    #include<cmath>
    #include<iomanip>
    #define INF 99999999
    using namespace std;
    
    const int MAX=100+10;
    double s[MAX][2];
    
    double calculate(int i,int j,int k){
    	return fabs((s[j][0]-s[i][0])*(s[k][1]-s[i][1])-(s[k][0]-s[i][0])*(s[j][1]-s[i][1]))/2;
    }
    
    int main(){
    	int t,n;
    	scanf("%d",&t);
    	while(t--){
    		scanf("%d",&n);
    		for(int i=0;i<n;++i)scanf("%lf%lf",&s[i][0],&s[i][1]);
    		double sum=INF*1.0;
    		for(int i=0;i<n;++i){
    			for(int j=i+1;j<n;++j){
    				for(int k=j+1;k<n;++k){
    					if(s[i][1] == s[j][1] && s[j][1] == s[k][1])continue;
    					if((s[j][1]-s[i][1])/(s[j][0]-s[i][0]) == (s[k][1]-s[j][1])/(s[k][0]-s[j][0]))continue;
    					sum=min(sum,calculate(i,j,k));
    				}
    			}
    		}
    		if(sum == INF*1.0)printf("Impossible
    ");
    		else printf("%.2lf
    ",sum);
    	}
    	return 0;
    }
  • 相关阅读:
    spring boot 扫描不到自定义Controller
    SpringBoot+Maven多模块项目(创建、依赖、打包可执行jar包部署测试)完整流程
    spring boot 中使用 jpa以及jpa介绍
    java8 快速实现List转map 、分组、过滤等操作
    Mysql 创建函数出现This function has none of DETERMINISTIC, NO SQL, or READS SQL DATA
    Spring mvc @initBinder 类型转化器的使用
    @RequestMapping 和@ResponseBody 和 @RequestBody和@PathVariable 注解 注解用法
    ssm的自动类型转换器
    如果将get请求转换成post请求
    如何将post请求转换成put和delete请求
  • 原文地址:https://www.cnblogs.com/james1207/p/3310559.html
Copyright © 2011-2022 走看看