Hdu4331 Image Recognition

Image Recognition

Time Limit: 6000/3000 MS (Java/others) memory limit: 65536/32768 K (Java/Others)
Total submission (s): 964 accepted submission (s): 371


Problem descriptionnow there is an image recognition problem for you. now you are given an image which is a n * n matrix and there are only 0 s and 1 s in the matrix. and we are interested in the squares in whose four edges there is no 0 s. so it's your task to find how many such
Squares in the image.

Inputthe first line of the input contains an integer T (1 <= T <= 10) which means the number of test cases.
For each test cases, the first line is one integer N (1 <= n <= 1000) which is the size of the image. then there are n lines and each line has n integers each of which is either 0 or 1.

Outputfor each test case, please output a line which is "case X: Y ", X means the number of the test case and Y means the number of the squares we are interested in the image.

Sample Input

131 1 01 1 00 0 0

 

Sample output

Case 1: 5

 

Source2012 multi-university training contest 4

Recommendzhoujiaqi2010 should be a line segment tree. I didn't expect it to be violent.

#include <iostream>#include <stdio.h>#include <string.h>using namespace std;#define MAXN 1005int up[MAXN][MAXN];int lleft[MAXN][MAXN];int map[MAXN][MAXN];int fmin(int a,int b){    if(a<b)    return a;    return b;}int main(){    int n,tcase,i,j,t=1;    int ans;    scanf("%d",&tcase);    while(tcase--)    {        scanf("%d",&n);        ans=0;        memset(lleft,0,sizeof(lleft));        memset(up,0,sizeof(up));        for(i=1;i<=n;i++)            for(j=1;j<=n;j++)            {                scanf("%d",&map[i][j]);                if(map[i][j])                {                    ans++;                    lleft[i][j]=lleft[i][j-1]+1;                    up[i][j]=up[i-1][j]+1;                }                else                {                    lleft[i][j]=0;                    up[i][j]=0;                }            }        for(i=1;i<=n;i++)        {            for(j=1;j<=n;j++)            {                              int k=fmin(up[i][j],lleft[i][j]);                if(k<=1)                continue;                int temp;                for(temp=2;temp<=k;temp++)                {                    if(up[i][j-temp+1]>=temp&&lleft[i-temp+1][j]>=temp)                    {                        ans++;                                       }                }            }                  }        printf("Case %d: %d\n",t++,ans);    }    return 0;}