bzoj 1132 POI2008 Tro

    大水题=_=，可我想复杂了……

    很裸的暴力，就是加了个小优化……

    叉积求面积 ：abs（xi*yj - yi*xj） 所以去掉绝对值，把 xi 和 xj 提出来就可以求和了

    去绝对值加个极角排序，每次把最左边的点当成原点，然后剩下的排序，接着枚举第二个点，求叉积之和……

    坐标都是整数，用long long，最后再除2

    上代码：

#include <cstdio>

#include <cstring>

#include <cstdlib>

#include <iostream>

#include <algorithm>

#include <cmath>

#define N 3010

using namespace std;



struct sss

{

    long long x, y;

}dian[N], now, zan[N];

int n;

long long ans = 0;



long long chaji(sss x, sss y)

{

    return (x.x-now.x)*(y.y-now.y) - (x.y-now.y)*(y.x-now.x);

}



bool cmp1(sss x, sss y) { return x.x == y.x ? x.y < y.y : x.x < y.x; }

bool cmp2(sss x, sss y ){ return chaji(x, y) > 0; }



int main()

{

    scanf("%d", &n);

    for (int i = 1; i <= n; ++i) scanf("%lld%lld", &dian[i].x, &dian[i].y);

    sort(dian+1, dian+1+n, cmp1);

    for (int i = 1; i <= n-2; ++i)

    {

        now = dian[i];

        long long ty = 0, tx = 0;

        for (int j = i+1; j <= n; ++j) zan[j] = dian[j];

        sort(zan+i+1, zan+1+n, cmp2);

        for (int j = i+1; j <= n; ++j)

        {

            ty += zan[j].y-now.y;

            tx += zan[j].x-now.x;

        }

        for (int j = i+1; j <= n-1; ++j)

        {

            ty -= zan[j].y-now.y; tx -= zan[j].x-now.x;

            ans += (zan[j].x-now.x)*ty - (zan[j].y-now.y)*tx;

        }

    }

    if (ans % 2) printf("%lld.5\n", ans/2);

    else printf("%lld.0\n", ans/2);

}