[luogu1034] 矩形覆盖 (暴力)

传送门

Description

给n(n<=50)个点(x,y)，要求用k(1<=k<=4)个没有联系的矩形覆盖住求矩形最小面积

Solution

感觉不是很可做，结果看TJ后发现数据太水直接暴力233

Code

//By Menteur_Hxy
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#define F(i,a,b) for(register int i=(a);i<=(b);i++)
using namespace std;

inline int read() {
	int x=0,f=1; char c=getchar();
	while(!isdigit(c)) {if(c=='-')f=-f;c=getchar();}
	while(isdigit(c)) x=(x<<1)+(x<<3)+c-48,c=getchar();
	return x*f;
}

const int N=60,INF=0x3f3f3f3f;
int n,k,ans;
int col[N],l[5],r[5],u[5],d[5],da[5];
struct Poi{int x,y;}p[N];

inline int get(int x) {
	int res=0;
	F(i,1,k) l[i]=u[i]=INF,r[i]=d[i]=da[i]=0;
	F(i,1,x) {
		int c=col[i];
		l[c]=min(l[c],p[i].y); u[c]=min(u[c],p[i].x);
		r[c]=max(r[c],p[i].y); d[c]=max(d[c],p[i].x);
		da[c]=(r[c]-l[c])*(d[c]-u[c]);
	}
	F(i,1,k) res+=da[i];
	return res;
}

inline void dfs(int x) {
	if(x==n+1) {
		int res=get(n);
		F(i,1,n) F(j,1,k) if(col[i]!=j)
			if(l[j]<=p[i].y&&p[i].y<=r[j]&&
			u[j]<=p[i].x&&p[i].x<=d[j]) return ;
		ans=min(ans,res);
		return ;
	}
	F(i,1,k) {
		col[x]=i;
		if(get(x)>ans) continue;
//		printf("%d %d %d %d %d\n",x,l[i],r[i],u[i],d[i]);
		dfs(x+1);
		col[x]=0;
	}
	return ;
}

int main() {
	n=read(),k=read();
	F(i,1,n) p[i].x=read(),p[i].y=read();
	ans=INF; dfs(1);
	printf("%d",ans);
	return 0;
}