【bzoj 4581】[Usaco2016 Open]Field Reduction（dfs）

4581: [Usaco2016 Open]Field Reduction

Time Limit: 10 Sec  Memory Limit: 128 MB
Submit: 84  Solved: 17
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Description

Farmer John's N cows (5≤N≤50,000) are all located at distinct positions in his two-dimensional fie

ld. FJ wants to enclose all of the cows with a rectangular fence whose sides are parallel to the x a

nd y axes, and he wants this fence to be as small as possible so that it contains every cow (cows on

 the boundary are allowed).FJ is unfortunately on a tight budget due to low milk production last qua

rter. He would therefore like to build an even smaller fenced enclosure if possible, and he is willi

ng to sell up to three cows from his herd to make this possible.Please help FJ compute the smallest 

possible area he can enclose with his fence after removing up to three cows from his herd (and there

after building the tightest enclosing fence for the remaining cows).For this problem, please treat c

ows as points and the fence as a collection of four line segments (i.e., don't think of the cows as 

"unit squares"). Note that the answer can be zero, for example if all remaining cows end up standing

 in a common vertical or horizontal line.

给定平面上N个点。现你可以删去至多3个点，接着你需要用一个矩形包含所有的点，点可以在矩形的边上，矩形的

边须与坐标轴平行。最小化矩形的面积并输出这个值。

5 ≤ N ≤ 50000, 1 ≤ X_i, Y_i ≤ 40000

Input

The first line of input contains N.

The next N lines each contain two integers specifying the locati

on of a cow. Cow locations are positive integers in the range 1…40,000.

Output

 Write a single integer specifying the minimum area FJ can enclose with his fence after removing up t

o three carefully-chosen cows from his herd.

Sample Input

6
1 1
7 8
10 9
8 12
4 100
50 7

Sample Output

12

HINT

Source

Silver鸣谢frank_c1提供翻译

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【题解】【dfs】

【大模拟，dfs暴力枚举，从四周去点，每次把一个边界的所有点都去掉。】

#include<cstdio>
#include<cstring>
#include<algorithm>
#define ll long long
#define inf 1e18
using namespace std;
struct node{
    ll heng,line;
}d[50010];
int n;
ll lmin,rmax,hmax,umin,ans;
bool vis[50010];
void dfs(int t)
{
	ll xmin,xmax,ymin,ymax;
	xmin=ymin=inf; xmax=ymax=-inf;
	for(int i=1;i<=n;++i)
	 if(!vis[i]) xmin=min(xmin,d[i].heng),xmax=max(xmax,d[i].heng),ymin=min(ymin,d[i].line),ymax=max(ymax,d[i].line);
	ll sum=(xmax-xmin)*(ymax-ymin);
	if(!t) 
	 {
	 	if(sum<ans) ans=sum;
	 	return;
	 }
	int cnt=0;
	for(int i=1;i<=n;++i)
	 if(!vis[i])
	  if(d[i].heng==xmin) cnt++,vis[i]=1;
	if(cnt&&cnt<=t)	dfs(t-cnt);
	cnt=0;
	for(int i=1;i<=n;++i)
	 {
	  	if(d[i].heng==xmin) vis[i]=0;
	  	if(d[i].heng==xmax) vis[i]=1,cnt++;
	 }
	if(cnt&&cnt<=t) dfs(t-cnt);
	cnt=0;
	for(int i=1;i<=n;++i)
	 {
	 	if(d[i].heng==xmax) vis[i]=0;
	 	if(d[i].line==ymin) vis[i]=1,cnt++;
	 }
	if(cnt&&cnt<=t) dfs(t-cnt);
	cnt=0;
	for(int i=1;i<=n;++i)
	 {
	 	if(d[i].line==ymin) vis[i]=0;
	 	if(d[i].line==ymax) vis[i]=1,cnt++;
	 }
	if(cnt&&cnt<=t) dfs(t-cnt);
	cnt=0;
	for(int i=1;i<=n;++i)
	 if(d[i].line==ymax) vis[i]=0;
}
int main()
{
//	freopen("reduce.in","r",stdin);
//	freopen("reduce.out","w",stdout);
	int i,j;
	scanf("%d",&n);
	lmin=umin=inf; rmax=hmax=-inf;
	for(i=1;i<=n;++i)
	 {
	 	ll x,y;
	 	scanf("%lld%lld",&x,&y);
        d[i].heng=x; d[i].line=y;
		if(lmin>x) lmin=x; if(rmax<x) rmax=x;	 
		if(hmax<y) hmax=y; if(umin>y) umin=y;	
	 }
	ans=(rmax-lmin)*(hmax-umin);
	dfs(3);
	printf("%lld\n",ans);
	return 0;
}

[96s跑进第一页，莫名开心啊……]