POJ 1654（计算几何基础多边形面积）

题意：给你一串数字，每个数字分别代表不同的方向，一定是在1x1的格子中走动，问最后围成的多边形面积是多少

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题解：将整个多边形划分，
ans = （∑相邻两个点分别与原点构成的线段的叉积 ）/ 2
由叉积的几何意义可知求得的结果是一个平行四边形的面积

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#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <string>#include <map>#include <cmath>#include <queue>#include <set>#define INF 0x3f3f3f3fusing namespace std;const int maxn = 1e6 + 5;typedef long long LL;int dx[] = {0,-1,0,1,-1,0,1,-1,0,1};int dy[] = {0,-1,-1,-1,0,0,0,1,1,1};struct Point{    int x,y;    Point(){}    Point(int xx, int yy):x(xx),y(yy){}};LL cross(Point a, Point b, Point c){    return (c.x-a.x)*(c.y-b.y)-(c.x-b.x)*(c.y-a.y);}int main(){    int n;    char s[maxn];    scanf("%d",&n);    while(n--){        scanf("%s",s);        int len = (int)strlen(s);        Point a(0,0);        Point dot(0,0);        LL ans = 0;        for(int i=0; i<len; i++){            Point b(a.x+dx[s[i]-'0'],a.y+dy[s[i]-'0']);            ans += cross(a,b,dot);            a = b;        }        if(ans < 0) ans = -ans;        printf("%lld",ans/2);        if(ans % 2 == 1) printf(".5");        printf("\n");    }}