hdu 01 Matrix

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 605    Accepted Submission(s): 127

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<n, m<=1000), specifying the size of matrix and the query number.
Then n lines follow and each line contains n chars ("0" or "1").
Then m lines follow, each lines contains a number k (0<k<=n).

 

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

 

这题用二维树状数组T了，改成用dp做，371ms过了。我们枚举每一个点作为右下角的点对各部分的贡献，那么就可以dp了。

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<string>
#include<algorithm>
using namespace std;
typedef long long ll;
#define pi acos(-1.0)
#define maxn 1006
int n;
char s[maxn][maxn];
int b[maxn],dp[maxn][maxn],heng[maxn][maxn],shu[maxn][maxn];
int lowbit(int x){
    return x&(-x);
}
void update(int pos,int num)
{
    while(pos<=maxn){
        b[pos]+=num;pos+=lowbit(pos);
    }
}
int getsum(int pos)
{
    int num=0;
    while(pos>0){
        num+=b[pos];pos-=lowbit(pos);
    }
    return num;
}


int main()
{
    int m,i,j,T,d;
    int l,r,mid;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        for(i=1;i<=n;i++){
            scanf("%s",s[i]+1);
        }
        for(i=1;i<=n;i++){
            b[i]=0;
            for(j=1;j<=n;j++){
                shu[i][j]=0;
                heng[i][j]=0;
            }
        }
        /*
        memset(b,0,sizeof(b));
        memset(shu,0,sizeof(shu));
        memset(heng,0,sizeof(heng));
        memset(dp,0,sizeof(dp));
        */
        for(i=1;i<=n;i++){
            for(j=1;j<=n;j++){
                if(i==1 && j==1){
                    if(s[i][j]=='1'){
                        dp[i][j]=shu[i][j]=heng[i][j]=1;
                        update(1,1);
                        update(2,-1);
                    }
                    else{
                        dp[i][j]=shu[i][j]=heng[i][j]=0;
                    }
                    continue;
                }
                if(s[i][j]=='0'){
                    dp[i][j]=shu[i][j]=heng[i][j]=0;
                    continue;
                }
                if(i==1){
                    dp[i][j]=1;
                    heng[i][j]=heng[i][j-1]+1;
                    shu[i][j]=1;
                    update(1,1);
                    update(dp[i][j]+1,-1);

                }
                else if(j==1){
                    dp[i][j]=1;
                    heng[i][j]=1;
                    shu[i][j]=shu[i-1][j]+1;
                    update(1,1);
                    update(dp[i][j]+1,-1);
                }
                else{
                    heng[i][j]=heng[i][j-1]+1;
                    shu[i][j]=shu[i-1][j]+1;
                    int len=min(heng[i][j],shu[i][j]);
                    len=min(len,dp[i-1][j-1]+1);
                    dp[i][j]=len;
                    update(1,1);
                    update(len+1,-1);
                }
            }
        }
        for(i=1;i<=m;i++){
            scanf("%d",&d);
            printf("%d
",getsum(d));
        }
    }
    return 0;
}