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\n",getsum(d));
}
}
return 0;
}