P1034 矩形覆盖

第一反应就是大法师：枚举横着切和竖着切。

感觉这样就已经对了，但后来想想貌似存在这样一种状况。

其实想想这种情况也不会是最优的。但是这题k等于四，总觉得不讨论一下就很不舒服。

然后就A掉了。

#include<cstdio>#include<iostream>#include<algorithm>#include<cstring>using namespace std;#define rep(i,j,k) for(i=j;i<=k;++i)#define per(i,j,k) for(i=j;i>=k;--i)#define ll long long#define db double#define mkp(x,y) make_pair(x,y)#define pii pair<int,int>#define X first#define Y secondconst int N=51,M=500;int n,k,ans;pii v[N];bool cmp(pii x,pii y){return x.Y==y.Y?x.X<y.X:x.Y<y.Y;}int cal(int l,int r,int d,int u){return (r-l)*(u-d);}void f1(pii v1,pii v2){int h,u1=v1.Y,l1=v1.X,d2=v2.Y,r2=v2.X,u3=0,d3=M,l3=M,r3=0,u4=0,d4=M,l4=M,r4=0;rep(h,1,n)if(v[h]!=v1&&v[h]!=v2){if(v[h].X<=v1.X&&v[h].Y>=v1.Y)u1=max(u1,v[h].Y),l1=min(l1,v[h].X);if(v[h].X>=v2.X&&v[h].Y<=v2.Y)d2=min(d2,v[h].Y),r2=max(r2,v[h].X);if(v[h].X>v1.X&&v[h].Y>v2.Y)u3=max(u3,v[h].Y),d3=min(d3,v[h].Y),l3=min(l3,v[h].X),r3=max(r3,v[h].X);if(v[h].X<v2.X&&v[h].Y<v1.Y)u4=max(u4,v[h].Y),d4=min(d4,v[h].Y),l4=min(l4,v[h].X),r4=max(r4,v[h].X);}ans=min(ans,cal(l1,v1.X,v1.Y,u1)+cal(v2.X,r2,d2,v2.Y)+cal(l3,r3,d3,u3)+cal(l4,r4,d4,u4));}void f2(pii v1,pii v2){int h,u1=v1.Y,r1=v1.X,d2=v2.Y,l2=v2.X,u3=0,d3=M,l3=M,r3=0,u4=0,d4=M,l4=M,r4=0;rep(h,1,n)if(v[h]!=v1&&v[h]!=v2){if(v[h].X>=v1.X&&v[h].Y>=v1.Y)u1=max(u1,v[h].Y),r1=max(r1,v[h].X);if(v[h].X<=v2.X&&v[h].Y<=v2.Y)d2=min(d2,v[h].Y),l2=max(l2,v[h].X);if(v[h].X<v1.X&&v[h].Y>v2.Y)u3=max(u3,v[h].Y),d3=min(d3,v[h].Y),l3=min(l3,v[h].X),r3=max(r3,v[h].X);if(v[h].X>v2.X&&v[h].Y<v1.Y)u4=max(u4,v[h].Y),d4=min(d4,v[h].Y),l4=min(l4,v[h].X),r4=max(r4,v[h].X);}ans=min(ans,cal(v1.X,r1,v1.Y,u1)+cal(l2,v2.X,d2,v2.Y)+cal(l3,r3,d3,u3)+cal(l4,r4,d4,u4));}int F1(int le,int ri,int cur);int F2(int le,int ri,int cur); int main(){int i,j,h,x,y;pii v1,v2;scanf("%d%d",&n,&k);rep(i,1,n){scanf("%d%d",&x,&y);v[i]=mkp(x,y);}sort(v+1,v+n+1);j=0;rep(i,1,n)if(v[i]!=v[j])v[++j]=v[i];n=j;if(k>=n){puts("0");return 0;}ans=M*M;if(k==4){per(i,n-1,1){rep(j,i+1,n){v1=v[i],v2=v[j];if(v1.X==v2.X)f1(v1,v2),f2(v1,v2);else if(v1.Y==v2.Y)f1(v1,v2),f2(v2,v1);else{x=min(v1.Y,v2.Y),y=max(v1.Y,v2.Y);rep(h,1,n)if(h!=i&&h!=j&&v[h].X>=v1.X&&v[h].X<=v2.X&&v[h].Y>=x&&v[h].Y<=y)break;if(h>n)if(v1.Y<v2.Y)f1(v2,v1),f1(v1,v2);else f2(v1,v2),f2(v2,v1);}}}}ans=min(ans,F1(1,n,k));printf("%d\n",ans);return 0;}int F1(int le,int ri,int cur){if(ri-le<cur)return 0;if(cur==1){int i,u=0,d=M,l=M,r=0;rep(i,le,ri)u=max(v[i].Y,u),d=min(v[i].Y,d),l=min(v[i].X,l),r=max(v[i].X,r);return cal(l,r,d,u);}int i,j,rtn=M*M;per(i,ri-1,le)per(j,cur-1,1)rtn=min(rtn,F1(le,i,j)+F1(i+1,ri,cur-j));sort(v+le,v+ri+1,cmp);per(i,ri-1,le)per(j,cur-1,1)rtn=min(rtn,F2(le,i,j)+F2(i+1,ri,cur-j));sort(v+le,v+ri+1);return rtn;}int F2(int le,int ri,int cur){if(ri-le<cur)return 0;if(cur==1){int i,u=0,d=M,l=M,r=0;rep(i,le,ri)u=max(v[i].Y,u),d=min(v[i].Y,d),l=min(v[i].X,l),r=max(v[i].X,r);return cal(l,r,d,u);}int i,j,rtn=M*M;per(i,ri-1,le)per(j,cur-1,1)rtn=min(rtn,F2(le,i,j)+F2(i+1,ri,cur-j));sort(v+le,v+ri+1);per(i,ri-1,le)per(j,cur-1,1)rtn=min(rtn,F2(le,i,j)+F2(i+1,ri,cur-j));sort(v+le,v+ri+1,cmp);return rtn;}