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 = 50010;
const int M = 1e6+10;
const int MOD = 1e9+7;
#define LL long long
#define LB long double
#define mi() (l+r)>>1
double const pi = acos(-1);
const double eps = 1e-8;
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_=0,int y_=0):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+1,p+n+1);
    for(int i=1;i<=n;i++){
        l[i]=min(l[i-1],p[i].y),r[i]=max(r[i-1],p[i].y);
    }
    for(int i=n;i>=1;i--){
        l1[i]=min(l1[i+1],p[i].y),r1[i]=max(r1[i+1],p[i].y);
    }
    for(int i=2;i<=n;i++){
        LL tem=(LL)(r[i-1]-l[i-1])*(p[i-1].x-p[1].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=1;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=1;i<=n;i++) swap(p[i].x,p[i].y);
    work();
    printf("%lld
",ans1-ans);
    return  0;
}