POJ - 1654 Area （计算几何 叉积 求多边形面积）

Area

 

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

题目链接：http://poj.org/problem?id=1654

题目大意：8,2,6,4 分别表示北，南，东，西， 9,7,3,1 分别表示东北，西北，东南，西南， 5表示结束动，起点（0,0），按照给的序列从起点开始画图，保证没有交叉的线，求所画图的面积

思路：用叉积求面积

代码：

#include
#include
#include
#include
using namespace std;
#define ll long long
const int N=1000005;
char ss[N];
struct node
{
    ll x,y;
};
ll go[][2]={-1,-1,0,-1,1,-1,-1,0,0,0,1,0,-1,1,0,1,1,1};
ll cross(node a,node b,node c)
{
    return (b.x-a.x)*(c.y-a.y)-(b.y-a.y)*(c.x-a.x);
}

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%s",ss);
        int n=strlen(ss);
        ll ans=0;
        node a,b,c;
        a.x=0,a.y=0;
        int k=ss[0]-'0'-1;
        b.x=go[k][0],b.y=go[k][1];
        for(int i=1;i