poj 1654 Area（求多边形面积）

题意：从原点出发向八个方向走，所给数字串每个数字代表一个方向，终点与原点连线，求所得多边形面积；

思路：（性质）共起点的两向量叉积的一半为两向量围成三角形的面积。以此计算每条边首尾两个向量的叉积，求和，除二；

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
const double epsi=1e-10;
const double pi=acos(-1.0);
const int maxn=100005;
inline int sign(const double &x){
    if(x>epsi) return 1;
    if(x<-epsi) return -1;
    return 0;
}
struct point{
    long long x,y;
    point(double xx,double yy):x(xx),y(yy) {}
    point operator +(const point &op2) const{
        return point(x+op2.x,y+op2.y);
    }
    long long operator ^(const point &op2) const{
        return x*op2.y-y*op2.x;
    }
};
int main()
{
    int t,i,j,k;
    string s;
    long long ans;
    scanf("%d",&t);
    while(t--){
        cin>>s;
        ans=0;
        point p=point(0,0),p1=point(0,0);
        for(int i=0;i<s.size();i++){
            if(s[i]=='1') p1=p+point(-1,-1);
            if(s[i]=='2') p1=p+point(0,-1);
            if(s[i]=='3') p1=p+point(1,-1);
            if(s[i]=='4') p1=p+point(-1,0);
            if(s[i]=='5') p1=p+point(0,0);
            if(s[i]=='6') p1=p+point(1,0);
            if(s[i]=='7') p1=p+point(-1,1);
            if(s[i]=='8') p1=p+point(0,1);
            if(s[i]=='9') p1=p+point(1,1);
            ans+=p^p1;
            p=p1;
        }
        if(ans<0) ans=-ans;
        printf("%lld",ans/2);
        if(ans%2) printf(".5");
        printf("
");
    }
    return 0;
}