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;
}
