SPOJ MINSUB 二分+单调栈

Largest Submatrix

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You are given an matrix M (consisting of nonnegative integers) and an integer K.  For any submatrix of M' of M define min(M') to be the minimum value of all the entries of M'.  Now your task is simple:  find the maximum value of min(M') where M' is a submatrix of M of area at least K (where the area of a submatrix is equal to the number of rows times the number of columns it has).

Input

The first line contains a single integer T (T ≤ 10) denoting the number of test cases, T test cases follow.  Each test case starts with a line containing three integers, R (R ≤ 1000), C (C ≤ 1000) and K (K ≤ R * C) which represent the number of rows, columns of the matrix and the parameter K.  Then follow R lines each containing C nonnegative integers, representing the elements of the matrix M.  Each element of M is ≤ 10^9

Output

For each test case output two integers:  the maximum value of min(M'), where M' is a submatrix of M of area at least K, and the maximum area of a submatrix which attains the maximum value of min(M').  Output a single space between the two integers.

Example

Input:22 2 21 11 13 3 21 2 34 5 67 8 9Output:1 48 2

题意：给你一个矩阵  找出一个子矩阵  面积至少为k的情况下里面的最小值最大

题解：二分答案  对于每次二分  创造出一个新的01矩阵  1代表当前为比此答案大  0代表小  问题就转化成了最大01子矩阵

用单调栈求解子矩阵

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int n,m,k;int a[1005][1005],b[1005][1005],d[1005],lef[1005][1005],rig[1005][1005];int solve(int x) {    int i,j,ans;ans=0;      for(j=1;j<=m;j++)b[1][j]=(a[1][j]>=x);      for(i=2;i<=n;i++)for(j=1;j<=m;j++)b[i][j]=(a[i][j]>=x)?(1+b[i-1][j]):0;      for(i=1;i<=n;i++)      {      d[0]=1;        d[1]=lef[i][1]=1;          for(j=2;j<=m;j++)          {              while(d[0]>0&&b[i][d[d[0]]]>=b[i][j])d[0]--;              lef[i][j]=(d[0])?(d[d[0]]+1):1;d[++d[0]]=j;          }          d[d[0]=1]=rig[i][m]=m;          for(j=m-1;j>=1;j--)          {              while(d[0]>0&&b[i][d[d[0]]]>=b[i][j])d[0]--;              rig[i][j]=(d[0])?(d[d[0]]-1):m;d[++d[0]]=j;          }          for(j=1;j<=m;j++)if(b[i][j])ans=max(ans,b[i][j]*(rig[i][j]-lef[i][j]+1));      }      return (ans>=k)?ans:0;  }  int main(){int t,i,j;scanf("%d",&t);while(t--){scanf("%d%d%d",&n,&m,&k);for(i=1;i<=n;i++){for(j=1;j<=m;j++)scanf("%d",&a[i][j]);}int l=0,r=1000000000;while(l+1<r){int mid=(l+r)/2;if(solve(mid))l=mid;else r=mid;}if(solve(r))l=r;printf("%d %d\n",l,solve(l));}return 0;}