【hdu 2036】改革春风吹满地

【题目链接】:http://acm.hdu.edu.cn/showproblem.php?pid=2036

【题意】

中文题

【题解】

这里用的是叉积对应的求三角形的面积;
即
A×B=A*B*sina
除2的话就能和面积对应了;
且因为算的是“有向面积”
所以就算是凹多边形也能正确计算;
叉积用行列式来记.

A×B=|i j k|
    |a b c|
    |d e f|
//在二维上就对应c和f为0的情况所以也即a*e-b*d

【Number Of WA】

0

【完整代码】

#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define ms(x,y) memset(x,y,sizeof x)

typedef pair<int,int> pii;
typedef pair<LL,LL> pll;

const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0);
const int N = 110;

struct point
{
    double x,y;
};

double chaji(point a,point b)
{
    double x1 = a.x,y1 = a.y;
    double x2 = b.x,y2 = b.y;
    return x1*y2-x2*y1;
}

int n;
point a[N];

int main()
{
    //freopen("F:\\rush.txt","r",stdin);
    ios::sync_with_stdio(false);
    while (cin >> n)
    {
        if (n==0) break;
        double x0,y0,x1,y1,x2,y2;
        cin >> x0 >> y0 >> x1 >> y1;
        double s = 0;
        rep1(i,3,n)
        {
            double x2,y2;
            cin >> x2>>y2;
            point t1,t2;
            t1.x=x1-x0,t1.y = y1-y0;
            t2.x=x2-x1,t2.y = y2-y1;
            s+=chaji(t1,t2);
            x1 = x2,y1 = y2;
        }
        s = fabs(s);
        s/=2.0;
        cout << fixed<<setprecision(1)<<s<<endl;
    }
    //printf("\n%.2lf sec \n", (double)clock() / CLOCKS_PER_SEC);
    return 0;
}