zoukankan      html  css  js  c++  java
  • poj 1654:Area 区域 ---- 叉积(求多边形面积)

    Area
     
    Time Limit: 1000MS   Memory Limit: 10000K
    Total Submissions: 19398   Accepted: 5311

    利用叉积求多边形面积,可以分解成多个三角形

    利用面积公式,S=1/2*abs((x0*y1-x1*y0)+(x1*y2-x2*y1)...+(xn*yn-1-xn-1*yn)+(xn*y0-x0*yn)) ,把相邻两点和原点组成一个三角形,而总面积是这n个三角形面积的和,而三角形面积是两个相邻边向量的叉积

    Description

    You are going to compute the area of a special kind of polygon. One vertex of the polygon is the origin of the orthogonal coordinate system. From this vertex, you may go step by step to the following vertexes of the polygon until back to the initial vertex. For each step you may go North, West, South or East with step length of 1 unit, or go Northwest, Northeast, Southwest or Southeast with step length of square root of 2. 
    你要计算一种特殊的多边形的面积。一个顶点的多边形是正交坐标系的原点。从这个顶点,你可以走了一步一步的多边形的顶点后,直到回到最初的顶点。为每个步骤你可以去北、西、南、东与步长为1单位,或去西北,东北,西南或东南√2步长。

    For example, this is a legal polygon to be computed and its area is 2.5: 

    Input

    The first line of input is an integer t (1 <= t <= 20), the number of the test polygons. Each of the following lines contains a string composed of digits 1-9 describing how the polygon is formed by walking from the origin. Here 8, 2, 6 and 4 represent North, South, East and West, while 9, 7, 3 and 1 denote Northeast, Northwest, Southeast and Southwest respectively. Number 5 only appears at the end of the sequence indicating the stop of walking. You may assume that the input polygon is valid which means that the endpoint is always the start point and the sides of the polygon are not cross to each other.Each line may contain up to 1000000 digits.

    Output

    For each polygon, print its area on a single line.

    Sample Input

    4
    5
    825
    6725
    6244865

    Sample Output

    0
    0
    0.5
    2

    #include<cstdio>
    #define ll long long
    ll n,m,t,x1,x2,y1,y2,ans;
    int dx[10]={0,1,1,1,0,0,0,-1,-1,-1};
    int dy[10]={0,-1,0,1,-1,0,1,-1,0,1};
    ll abs(ll a){return a<0?-a:a;}
    int main()
    {
    	scanf("%d",&t);
    	for(int i=0;i<t;i++)
    	{
    		x2=y2=ans=0;
    		char p=getchar();
    		while(p=getchar())
    		{
    			ll tmp=p-'0';
    			if(tmp==5) break;
    			x1=x2,y1=y2;
    			x2+=dx[tmp],y2+=dy[tmp];
    			ans+=(x1*y2-x2*y1);
    		}
    		ans=abs(ans);
    		if(ans%2==0) printf("%lld
    ",ans/2);
    		else printf("%lld.5
    ",ans/2);
    	}
    }
    

      

    版权声明:本文为博主原创文章,未经博主允许不得转载。

  • 相关阅读:
    禁止在工作流设计器启动持续活动的重新编译
    设计流程 工作流
    workflow 工作流
    访问调度控制 时间控件
    如何:实现一个视图项目
    Python多线程之threading.Thread实现
    gcc 编译流程分析
    如何编写Makefile?
    linux 文件夹的颜色代表什么意思
    STL容器的迭代器失效的原因
  • 原文地址:https://www.cnblogs.com/lkhll/p/5978435.html
Copyright © 2011-2022 走看看