洛谷P1378 油滴扩展

题目链接

P1378 油滴扩展

解题思路：

#include <iostream>
#include<cstdio>
#include<cmath> 
#include<cstring>
#define PI 3.1415926535
using namespace std;
struct node{
	int x,y;
}a[10];
double maxv=0.0,b[10];
int vis[10];
int n;
int x,y,x2,y2;

double calc_r(double mx,double my,int i){
	double tx=min(abs(mx-x+0.0),abs(mx-x2+0.0));
	double ty=min(abs(my-y+0.0),abs(my-y2+0.0)); 
	double r=min(tx,ty);
	for(int j=1;j<=n;j++){
		if(i!=j&&vis[j]){//用vis记录 
			double dist=sqrt((mx-a[j].x)*(mx-a[j].x)+(my-a[j].y)*(my-a[j].y));
			if(dist-b[j]<=0)  return 0;//小于0，说明在已知的圆内，半径记为0 
			r=min(r,dist-b[j]);//dist-b[j]可能为负 
		}
	}
	return r;
}
void dfs(double sum,int step){
	if(step==n){
		maxv=max(maxv,sum);
		return;
	} 
	for(int i=1;i<=n;i++){//跟先后顺序有关 
		if(!vis[i]){
			vis[i]=1;
			b[i]=calc_r(a[i].x+0.0,a[i].y+0.0,i);
//			printf("%lf\n",b[i]);
			dfs(sum+PI*b[i]*b[i],step+1);
			vis[i]=0;
			b[i]=0.0;
		}
	}
}
int main(int argc, char** argv) {
	scanf("%d",&n);
	scanf("%d%d%d%d",&x,&y,&x2,&y2);
	for(int i=1;i<=n;i++)
		scanf("%d%d",&a[i].x,&a[i].y);

	dfs(0,0);
	double ans=abs(x-x2)*abs(y-y2)+0.0-maxv;
	printf("%.0lf\n",ans);
	return 0;
}