bzoj2228 礼物 单调队列

       首先枚举三个面作为底面。然后对于每一层，求出f[i][j]表示以i,j为右下角坐标的正方形的最大边长。然后枚举x,y表示以x,y为取出的长方体的正方形的底面，然后就变成了单调队列求区间长度*区间最小值的最大值的经典问题了。

AC代码如下（跑得好慢QAQ）：

#include<iostream>#include<cstdio>#include<cstring>#define N 151using namespace std;int n,m,w,a[N][N][N],b[N][N][N],f[N][N][N],c[N],h[N],l[N],r[N],ans;char s[N];void turn1(){int i,j,k;for (i=1; i<=m; i++)for (j=1; j<=n; j++)for (k=1; k<=w; k++) a[i][w-k+1][j]=b[i][j][k];swap(n,w);}void turn2(){int i,j,k; swap(n,w);for (i=1; i<=m; i++)for (j=1; j<=n; j++)for (k=1; k<=w; k++) a[k][j][m-i+1]=b[i][j][k];swap(m,w);}int calc(int cnt){int i,j=1,tmp=0; c[0]=c[cnt+1]=-1; h[1]=0;for (i=1; i<=cnt; i++){while (c[i]<=c[h[j]]) j--;l[i]=h[j]+1; h[++j]=i;}j=1; h[1]=cnt+1;for (i=cnt; i; i--){while (c[i]<=c[h[j]]) j--;r[i]=h[j]-1; h[++j]=i;}for (i=1; i<=cnt; i++) tmp=max(tmp,(r[i]-l[i]+1)*c[i]);return tmp;}void solve(){int i,j,k;for (k=1; k<=w; k++)for (i=1; i<=m; i++)for (j=1; j<=n; j++)if (a[i][j][k]) f[i][j][k]=min(min(f[i][j-1][k],f[i-1][j][k]),f[i-1][j-1][k])+1;else f[i][j][k]=0;for (i=1; i<=m; i++)for (j=1; j<=n; j++){memcpy(c,f[i][j],sizeof(f[i][j]));ans=max(ans,calc(w));}}int main(){scanf("%d%d%d",&m,&n,&w); int i,j,k;for (j=1; j<=n; j++)for (i=1; i<=m; i++){char ch=getchar(); while (ch<'A' || ch>'Z') ch=getchar();for (k=1; k<=w; k++){a[i][j][k]=b[i][j][k]=(ch=='N')?1:0; ch=getchar();}}solve(); turn1(); solve(); turn2();solve(); printf("%d\n",ans<<2);return 0;}

by lych

2016.2.26