bzoj4578: [Usaco2016 OPen]Splitting the Field

2365: Splitting the Field

题意：n个点，求用两个矩形面积覆盖完所有点和一个矩形覆盖完少多少面积

思路：枚举两个矩形的分割线，也就是把所有点分成两个部分，枚举分割点；先预处理每个点之前和之后的最大，最低高度；

矩形可以横着分，也可以竖着分

 // #pragma comment(linker, "/STACK:102c000000,102c000000")
 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <sstream>
 #include <string>
 #include <algorithm>
 #include <list>
 #include <map>
 #include <vector>
 #include <queue>
 #include <stack>
 #include <cmath>
 #include <cstdlib>
 // #include <conio.h>
 using namespace std;
 #define clc(a,b) memset(a,b,sizeof(a))
 #define inf 0x3f3f3f3f
 #define lson l,mid,rt<<1
 #define rson mid+1,r,rt<<1|1
 const int N = ;
 const int M = 1e6+;
 const int MOD = 1e9+;
 #define LL long long
 #define LB long double
 #define mi() (l+r)>>1
 double const pi = acos(-);
 const double eps = 1e-;
 void fre() {
     freopen("in.txt","r",stdin);
 }
 // inline int r() {
 //     int x=0,f=1;char ch=getchar();
 //     while(ch>'9'||ch<'0') {if(ch=='-') f=-1;ch=getchar();}
 //     while(ch>='0'&&ch<='9') { x=x*10+ch-'0';ch=getchar();}return x*f;
 // }
 struct point{
     int x,y;
     point(int x_=,int y_=):x(x_),y(y_){}
     bool operator < (const point &a) const{
         return x==a.x?y<a.y:x<a.x;
     }
 }p[N];
 int l[N],r[N],l1[N],r1[N];
 int n;
 LL ans;
 void work(){
     clc(l,inf),clc(r,-inf),clc(l1,inf),clc(r1,-inf);
     sort(p+,p+n+);
     for(int i=;i<=n;i++){
         l[i]=min(l[i-],p[i].y),r[i]=max(r[i-],p[i].y);
     }
     for(int i=n;i>=;i--){
         l1[i]=min(l1[i+],p[i].y),r1[i]=max(r1[i+],p[i].y);
     }
     for(int i=;i<=n;i++){
         LL tem=(LL)(r[i-]-l[i-])*(p[i-].x-p[].x)+(LL)(r1[i]-l1[i])*(p[n].x-p[i].x);
         ans=min(tem,ans);
     }
 }
 int main(){
     int minx=inf,miny=inf,maxx=-inf,maxy=-inf;
     scanf("%d",&n);
     for(int i=;i<=n;i++){
         scanf("%d%d",&p[i].x,&p[i].y);
         minx=min(minx,p[i].x);
         maxx=max(maxx,p[i].x);
         miny=min(miny,p[i].y);
         maxy=max(maxy,p[i].y);
     }
     ans=(LL)(maxx-minx)*(maxy-miny);
     LL ans1=ans;
     work();
     for(int i=;i<=n;i++) swap(p[i].x,p[i].y);
     work();
     printf("%lld\n",ans1-ans);
     return  ;
 }