2015‘12杭电校赛1006 01 Matrix（二维DP）

题解

我们用d[i][j]表示右下角为(i,j)的最大的全1正方形的size是多少。

那么直接有一个DP转移

if(s[i][j]==’1’)d[i][j]=min(d[i-1][j],min(d[i][j-1],d[i-1][j-1]))+1;

else d[i][j]=0;

然后一样计数++f[d[i][j]]

最后扫描一遍就可以啦。

代码

#include<cstdio>
#include<cstring>
#include<string>
#include<iostream>
#include<algorithm>
#include<map>
using namespace std;
const int N=50005;
int n,k,m;
int s[1005];
int v[1005][1005],r[1005][1005],x[1005][1005],a[1005][1005];
char c[1005];
int main()
{
 int i,j;
 int T;
 scanf("%d",&T);
 while(T--)
 {
 scanf("%d%d",&n,&m);
 memset(s,0,sizeof(s));
 memset(r,0,sizeof(r));
 memset(x,0,sizeof(x));
 for(i=1; i<=n; i++)
 {
 scanf("%s",c);
 for(j=0;j<n;j++)
 a[i][j+1]=c[j]-'0';
 }
 x[n+1][n]=0;
 r[n][n+1]=0;
 v[n+1][n+1]=0;
 for(i=n; i>=1; i--)
 {
 for(j=n; j>=1; j--)
 {
 if(a[i][j])
 {
 v[i][j]=min(min(v[i+1][j+1],r[i][j+1]),x[i+1][j]);
 r[i][j]=r[i][j+1]+1;
 x[i][j]=x[i+1][j]+1;
 v[i][j]++;
 }
 else
 v[i][j]=r[i][j]=x[i][j]=0;
 s[v[i][j]]++;
 }
 }
 for(i=n-1; i>=1; i--)
 {
 s[i]+=s[i+1];
 }
 while(m--)
 {
 scanf("%d",&k);
 cout<<s[k]<<endl;
 }
 }
 return 0;
}

题目

01 Matrix

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 666 Accepted Submission(s): 156

Problem Description

It’s really a simple problem.
Given a “01” matrix with size by n*n (the matrix size is n*n and only contain “0” or “1” in each grid), please count the number of “1” matrix with size by k*k (the matrix size is k*k and only contain “1” in each grid).

Input

There is an integer T (0 < T <=50) in the first line, indicating the case number.
Each test case begins with two numbers n and m (0

Output

For each query, output the number of “1” matrix with size by k*k.

Sample Input

2
2 2
01
00
1
2
3 3
010
111
111
1
2
2

Sample Output

1
0
7
2
2