BZOJ1069: [SCOI2007]最大土地面积

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传送门

对于给定的点，先求出凸包，听说水平序求凸包会被卡..亲测不会

然后对于求出来的凸包，求出每一个对踵点。然后对于每一个对踵点，遍历凸包上每一个点，求出最大的叉积和最小的叉积，绝对值的累加即位最大面积。

//BZOJ1069
//by Cydiater
//2017.1.29
#include <iostream>
#include <queue>
#include <map>
#include <cstring>
#include <string>
#include <cstdlib>
#include <cmath>
#include <ctime>
#include <iomanip>
#include <algorithm>
#include <cstdio>
#include <bitset>
#include <set>
#include <vector>
using namespace std;
#define ll long long
#define up(i,j,n)	for(int i=j;i<=n;i++)
#define down(i,j,n)	for(int i=j;i>=n;i--)
#define cmax(a,b)	a=max(a,b)
#define cmin(a,b)	a=min(a,b)
#define db 		double
#define Vector 		Point
const int MAXN=1e4+5;
const int oo=0x3f3f3f3f;
const db eps=1e-10;
struct Point{
	db x,y;
	Point(db x=0,db y=0):x(x),y(y){}
};
Vector operator + (Point x,Point y){return Vector(x.x+y.x,x.y+y.y);}
Vector operator - (Point x,Point y){return Vector(x.x-y.x,x.y-y.y);}
Vector operator * (Vector x,db p){return Vector(x.x*p,x.y*p);}
Vector operator / (Vector x,db p){return Vector(x.x/p,x.y/p);}
int dcmp(db x){if(fabs(x)<eps)return 0;else return x<0?-1:1;}
bool operator < (const Vector &x,const Vector &y){return dcmp(x.x-y.x)==0?x.y<y.y:x.x<y.x;}
bool operator == (const Vector &x,const Vector &y){return dcmp(x.x-y.x)==0&&dcmp(x.y-y.y)==0;}
int N,top=0;
Point V[MAXN],q[MAXN];
db ans=0;
namespace solution{
	Point Write(){db x,y;scanf("%lf%lf",&x,&y);return Point(x,y);}
	db Cross(Vector x,Vector y){return x.x*y.y-x.y*y.x;}
	void Prepare(){
		scanf("%d",&N);
		up(i,1,N)V[i]=Write();
		sort(V+1,V+N+1);
		N=unique(V+1,V+N+1)-(V+1);
	}
	void Andrew(){
		up(i,1,N){
			while(top>=2&&dcmp(Cross(q[top]-q[top-1],V[i]-q[top-1]))<=0)top--;
			q[++top]=V[i];
		}
		int lim=top;
		down(i,N-1,1){
			while(top-lim>=1&&dcmp(Cross(q[top]-q[top-1],V[i]-q[top-1]))<=0)top--;
			q[++top]=V[i];
		}
	}
	db Area(Point A,Point B){
		db area1=0,area2=0;
		up(i,1,top){
			cmax(area1,Cross(A-q[i],B-q[i])/2);
			cmin(area2,Cross(A-q[i],B-q[i])/2);
		}
		return area1-area2;
	}
	void Solve(){
		Andrew();
		top--;
		int pos=2;
		up(i,1,top){
			while(fabs(Cross(q[pos+1]-q[i+1],q[i]-q[i+1]))>fabs(Cross(q[pos]-q[i+1],q[i]-q[i+1]))){
				pos%=top;pos++;
			}
			cmax(ans,Area(q[pos],q[i]));
		}
		printf("%.3lf\n",ans);
	}
}
int main(){
	//freopen("input.in","r",stdin);
	using namespace solution;
	Prepare();
	Solve();
	return 0;
}

BZOJ1069: [SCOI2007]最大土地面积

标签：www   oid   ref   zoj   printf   scan   php   names   1.2   

原文地址：http://www.cnblogs.com/Cydiater/p/6357474.html