hdu大学生程序设计竞赛（2015’12）1006 01 Matrix

01 Matrix

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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

22 20100123 3010111111122

 

Sample Output

10722

 

/*题解:如果从左上角向右下角,每个点对应最大矩形的边长是很难考虑清楚地所以可以从右下角想左上角进行转移(dp是从局部到整体的思想)Down[i][j]表示这个点的下方有多少个连续的'1'Right[i][j]表示这个点的右方有多少个连续的'1'dp[i][j]表示以这个点为左上方顶点的矩形边长最大为多少状态转移方程:如果这个点对应的s[i][j]为'0',那么所有的数组均为0相反,则Down[i][j]=Down[i+1][j]+1,Right[i][j]=Right[i][j+1],dp[i][j]讨论一下就可以了然后知道了所有点的dp[i][j],利用后缀和的便可以求出每个边长有多少个方案了*/#include <map>#include <set>#include <stack>#include <queue>#include <cmath>#include <ctime>#include <vector>#include <cstdio>#include <cctype>#include <cstring>#include <cstdlib>#include <iostream>#include <algorithm>using namespace std;#define INF 0x3f3f3f3f#define inf -0x3f3f3f3f#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define mem0(a) memset(a,0,sizeof(a))#define mem1(a) memset(a,-1,sizeof(a))#define mem(a, b) memset(a, b, sizeof(a))typedef long long ll;char s[1102][1102];int Down[1102][1102],Right[1102][1102],dp[1102][1102];int C[1002];int main(){    int n,t,m;    scanf("%d",&t);    while(t--){        scanf("%d%d",&n,&m);        for(int i=1;i<=n;i++)            scanf("%s",s[i]+1);        for(int i=1;i<=n+10;i++){            for(int j=1;j<=n+10;j++)                dp[i][j]=0,Down[i][j]=0,Right[i][j]=0;            C[i]=0;        }        for(int i=n;i>=1;i--)            for(int j=n;j>=1;j--){                if(s[i][j]=='0'){                    Down[i][j]=0;                    dp[i][j]=0;                    Right[i][j]=0;                }                else{                    Right[i][j]=Right[i][j+1]+1;                    Down[i][j]=Down[i+1][j]+1;                    if(Right[i][j]>=dp[i+1][j+1]+1&&Down[i][j]>=dp[i+1][j+1]+1)                        dp[i][j]=dp[i+1][j+1]+1;                    else                        dp[i][j]=min(Down[i][j],Right[i][j]);                    C[dp[i][j]]++;                }            }        for(int i=n;i>=1;i--)            C[i]+=C[i+1];        while(m--){            int k;            scanf("%d",&k);            printf("%d\n",C[k]);        }    }    return 0;}