bzoj 4578 [Usaco2016 OPen]Splitting the Field

最新推荐文章于 2019-07-20 18:49:00 发布

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最新推荐文章于 2019-07-20 18:49:00 发布

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本文提供了一道USACO竞赛题目“Splitting the Field”的解决方案，该题要求用两个矩形最小化覆盖所有点的面积总和，并确保两个矩形不重叠。文章详细介绍了通过坐标排序和遍历来找到最优解的方法。

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4578: [Usaco2016 OPen]Splitting the Field

Time Limit: 10 Sec Memory Limit: 128 MB
Submit: 105 Solved: 46
[ Submit][ Status][ Discuss]

Description

Farmer John's N cows (3≤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 everycow (cows on

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

ter. He would therefore like to enclose a smaller area to reduce maintenance costs, and the only way

he can see to do this is by building two enclosures instead of one. Please help him compute how muc

h less area he needs to enclose, in total, by using two enclosures instead of one. Like the original

enclosure, the two enclosures must collectively contain all the cows (with cows on boundaries allow

ed), and they must have sides parallel to the x and y axes. The two enclosures are notallowed to ove

rlap -- not even on their boundaries. Note that enclosures of zero area are legal, for example if an

enclosure has zero width and/or zero height.

Input

The first line of input contains N. The nextNN lines each contain two integers specifying the location of a cow. Cow locations are positive integers in the range 1…1,000,000,000

Output

Write a single integer specifying amount of total area FJ can save by using two enclosures instead o

f one.

Sample Input

6
4 2
8 10
1 1
9 12
14 7
2 3

Sample Output

107

HINT

Source

Gold

【分析】

签到题...

大意：给二维坐标系中的n个点，求ans=用一个矩形覆盖所有点所用矩形面积-用两个矩形覆盖所有点所用两个矩形的最小面积和，而且两个矩形不能重合(边重合也不行)

手玩可知，两个矩形一定是一个在左边，一个在右边，或者一个在下面，一个在上面...分别x，y排序，然后扫一遍，同时记录坐标最大最小值就算一下就OK

【代码】

//bzoj 4578
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#define inf 1e9+7
#define ll long long
#define M(a) memset(a,0,sizeof a)
#define fo(i,j,k) for(i=j;i<=k;i++)
using namespace std;
const int mxn=50005;
ll ans=1e18+7,tot;
int n,m,mx,mn;
struct mnmx
{
	int minx[mxn],miny[mxn],maxx[mxn],maxy[mxn];
}A,B;
struct node {int x,y;} a[mxn];
inline bool comp_x(node a,node b)
{
	if(a.x==b.x) return a.y<b.y;
	return a.x<b.x;
}
inline bool comp_y(node a,node b)
{
	if(a.y==b.y) return a.x<b.x;
	return a.y<b.y;
}
inline int read()
{
	int x=0,f=1;char ch=getchar();
	while(ch<'0'||ch>'9') {if(ch=='-') f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9') x=(x<<1)+(x<<3)+ch-'0',ch=getchar();
	return x*f;
}
int main()
{
	int i,j;
	n=read();
	fo(i,1,n) a[i].x=read(),a[i].y=read();
	
	sort(a+1,a+n+1,comp_x);
	ll ox=a[n].x-a[1].x;
	fo(i,0,n+1) A.miny[i]=B.miny[i]=A.minx[i]=B.minx[i]=inf;
	fo(i,1,n) A.miny[i]=min(A.miny[i-1],a[i].y),A.maxy[i]=max(A.maxy[i-1],a[i].y);
	for(i=n;i>=1;i--) B.miny[i]=min(B.miny[i+1],a[i].y),B.maxy[i]=max(B.maxy[i+1],a[i].y);
	fo(i,1,n-1) ans=min(ans,(ll)(A.maxy[i]-A.miny[i])*(a[i].x-a[1].x)+(ll)(B.maxy[i+1]-B.miny[i+1])*(a[n].x-a[i+1].x));
	
	sort(a+1,a+n+1,comp_y);
	ll oy=a[n].y-a[1].y;
	fo(i,1,n) A.minx[i]=min(A.minx[i-1],a[i].x),A.maxx[i]=max(A.maxx[i-1],a[i].x);
	for(i=n;i>=1;i--) B.minx[i]=min(B.minx[i+1],a[i].x),B.maxx[i]=max(B.maxx[i+1],a[i].x);
	fo(i,1,n-1) ans=min(ans,(ll)(A.maxx[i]-A.minx[i])*(a[i].y-a[1].y)+(ll)(B.maxx[i+1]-B.minx[i+1])*(a[n].y-a[i+1].y));
	
//	printf("%lld\n",ans);
	printf("%lld\n",ox*oy-ans);
	return 0;
}