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
1 // #pragma comment(linker, "/STACK:102c000000,102c000000") 2 #include <iostream> 3 #include <cstdio> 4 #include <cstring> 5 #include <sstream> 6 #include <string> 7 #include <algorithm> 8 #include <list> 9 #include <map> 10 #include <vector> 11 #include <queue> 12 #include <stack> 13 #include <cmath> 14 #include <cstdlib> 15 // #include <conio.h> 16 using namespace std; 17 #define clc(a,b) memset(a,b,sizeof(a)) 18 #define inf 0x3f3f3f3f 19 #define lson l,mid,rt<<1 20 #define rson mid+1,r,rt<<1|1 21 const int N = 50010; 22 const int M = 1e6+10; 23 const int MOD = 1e9+7; 24 #define LL long long 25 #define LB long double 26 #define mi() (l+r)>>1 27 double const pi = acos(-1); 28 const double eps = 1e-8; 29 void fre() { 30 freopen("in.txt","r",stdin); 31 } 32 // inline int r() { 33 // int x=0,f=1;char ch=getchar(); 34 // while(ch>'9'||ch<'0') {if(ch=='-') f=-1;ch=getchar();} 35 // while(ch>='0'&&ch<='9') { x=x*10+ch-'0';ch=getchar();}return x*f; 36 // } 37 struct point{ 38 int x,y; 39 point(int x_=0,int y_=0):x(x_),y(y_){} 40 bool operator < (const point &a) const{ 41 return x==a.x?y<a.y:x<a.x; 42 } 43 }p[N]; 44 int l[N],r[N],l1[N],r1[N]; 45 int n; 46 LL ans; 47 void work(){ 48 clc(l,inf),clc(r,-inf),clc(l1,inf),clc(r1,-inf); 49 sort(p+1,p+n+1); 50 for(int i=1;i<=n;i++){ 51 l[i]=min(l[i-1],p[i].y),r[i]=max(r[i-1],p[i].y); 52 } 53 for(int i=n;i>=1;i--){ 54 l1[i]=min(l1[i+1],p[i].y),r1[i]=max(r1[i+1],p[i].y); 55 } 56 for(int i=2;i<=n;i++){ 57 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); 58 ans=min(tem,ans); 59 } 60 } 61 int main(){ 62 int minx=inf,miny=inf,maxx=-inf,maxy=-inf; 63 scanf("%d",&n); 64 for(int i=1;i<=n;i++){ 65 scanf("%d%d",&p[i].x,&p[i].y); 66 minx=min(minx,p[i].x); 67 maxx=max(maxx,p[i].x); 68 miny=min(miny,p[i].y); 69 maxy=max(maxy,p[i].y); 70 } 71 ans=(LL)(maxx-minx)*(maxy-miny); 72 LL ans1=ans; 73 work(); 74 for(int i=1;i<=n;i++) swap(p[i].x,p[i].y); 75 work(); 76 printf("%lld ",ans1-ans); 77 return 0; 78 }