2015'12항전교전 1006 01 Matrix(2D DP)

4378 단어

문제풀이


우리는 d[i][j]로 오른쪽 아래쪽이 (i, j)인 가장 큰 전1정사각형의 사이즈가 얼마인지 나타낸다.


그러면 바로 DP 마이그레이션이 있습니다.


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;


그리고 같은 계수 ++f [d[i][j]]


마지막으로 스캔하면 돼요.


코드


#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;
}

제목.


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

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