POJ 1654 Area

最新推荐文章于 2020-04-14 23:20:30 发布

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最新推荐文章于 2020-04-14 23:20:30 发布

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分类专栏：数学

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本文介绍了一种通过输入特定序列来计算特殊多边形面积的方法，并提供了完整的C++实现代码。该方法允许从原点出发，按指定方向行走形成多边形，最终计算其面积。

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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.

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

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一道waterful的叉积问题，然而还是wa了许多次。
1 需要注意是向量（有方向），所以中间不能取绝对值。
2 要用 long long。

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#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
long long ans;
char c[1000001];
int movx[10]={0,-1,0,1,-1,0,1,-1,0,1},movy[10]={0,-1,-1,-1,0,0,0,1,1,1};
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        ans=0;
        long long x1=0,y1=0,x2=0,y2=0;
        scanf("%s",c);
        int l=strlen(c);
        for(int i=0;i<=l-1;i++)
        {
            x1+=movx[c[i]-'0'];
            y1+=movy[c[i]-'0'];
            ans+=x1*y2-x2*y1;
            x2=x1;
            y2=y1;
        }
        if(ans<0)
            ans*=-1;
        printf("%lld",ans/2);
        if(ans&1)
            printf(".5");
        printf("\n");
    }
}