【刷题】BZOJ 1132 [POI2008]Tro

Description

平面上有N个点. 求出所有以这N个点为顶点的三角形的面积和 N<=3000

Input

第一行给出数字N,N在[3,3000] 下面N行给出N个点的坐标,其值在[0,10000]

Output

保留一位小数,误差不超过0.1

Sample Input

5

0 0

1 2

0 2

1 0

1 1

Sample Output

7.0

Solution

\(ans=\frac{1}{2}\sum_{i=1}^n\sum_{j=i+1}^n\sum_{k=j+1}^n|(y_j-y_i)(x_k-x_i)-(y_k-y_i)(x_j-x_i)|\)

枚举第一个点，求出其它点的相对坐标

然后为了去绝对值，让所有点按计较排序，保证叉积是正的

\(ans_i=\frac{1}{2}\sum_{j=i+1}^n\sum_{k=j+1}^ny_j*x_k-y_k*x_j\)

\(~~~~~~~~~=\frac{1}{2}(\sum_{j=i+1}^ny_j\sum_{k=j+1}^nx_k-\sum_{j=i+1}^nx_j\sum_{k=j+1}^ny_k)\)

对最后的 \(\sum\) 做前缀和就好了

#include<bits/stdc++.h>
#define ui unsigned int
#define ll long long
#define db double
#define ld long double
#define ull unsigned long long
#define REP(a,b,c) for(register int a=(b),a##end=(c);a<=a##end;++a)
#define DEP(a,b,c) for(register int a=(b),a##end=(c);a>=a##end;--a)
const int MAXN=3000+10;
int n,cnt;
ld ans;
struct point{
	int x,y;
	inline bool operator < (const point &A) const {
		return y<A.y;
	};
};
point pt[MAXN];
struct cross{
	int x,y;
	ld k;
	inline bool operator < (const cross &A) const {
		return k>A.k;
	};
};
cross cs[MAXN];
template<typename T> inline void read(T &x)
{
	T data=0,w=1;
	char ch=0;
	while(ch!='-'&&(ch<'0'||ch>'9'))ch=getchar();
	if(ch=='-')w=-1,ch=getchar();
	while(ch>='0'&&ch<='9')data=((T)data<<3)+((T)data<<1)+(ch^'0'),ch=getchar();
	x=data*w;
}
template<typename T> inline void write(T x,char ch='\0')
{
	if(x<0)putchar('-'),x=-x;
	if(x>9)write(x/10);
	putchar(x%10+'0');
	if(ch!='\0')putchar(ch);
}
template<typename T> inline void chkmin(T &x,T y){x=(y<x?y:x);}
template<typename T> inline void chkmax(T &x,T y){x=(y>x?y:x);}
template<typename T> inline T min(T x,T y){return x<y?x:y;}
template<typename T> inline T max(T x,T y){return x>y?x:y;}
int main()
{
	read(n);
	REP(i,1,n)read(pt[i].x),read(pt[i].y);
	std::sort(pt+1,pt+n+1);
	REP(i,3,n)
	{
		REP(j,1,i-1)cs[j]=(cross){pt[i].x-pt[j].x,pt[i].y-pt[j].y,atan2((ld)(pt[i].x-pt[j].x),(ld)(pt[i].y-pt[j].y))};
		std::sort(cs+1,cs+i);ld sx=0,sy=0;
		REP(j,1,i-1)
		{
			if(j!=1)ans+=sx*(ld)cs[j].y-(ld)cs[j].x*sy;
			sx+=(ld)cs[j].x,sy+=(ld)cs[j].y;
		}
	}
	printf("%.1Lf\n",ans/2);
	return 0;
}