CodeForces Gym 100935C OCR (水

OCR

Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

 

Description

standard input/output
Statements

Optical Character Recognition (OCR) is one of the most famous fields of Artificial Intelligence. The main purpose of OCR is to recognize printed text (or handwriting) and convert it to the machine encoded-text. You may have seen similar applications in your smartphone: you use your camera to take a photo that contains text, then, the text is translated or saved in PDF, for example. In this problem we deal with a very limited case of OCR. You have a scanned character which is either ‘0’ (number zero) or ‘8’ (number eight) and you have to decide what number it is. The input will be an image that contains exactly one character (It is guaranteed that this character is either ‘0’ or ‘8’). The image has 2 colors: white (represented with dot ‘.’) and black (represented with asterisk ‘*’). For simplicity, the borders of the image are always white. It’s also guaranteed that black lines inside the image are either vertical or horizontal. So you may safely assume that the shapes of ‘0’ and ‘8’ inside the image are the same as their shapes in digital clocks. However, they might be stretched or not positioned in the center of the image.

Input

The first line will be the number of test cases T (T<100). Each test case starts with two positive integers (n,m) denoting the dimensions of the image. (n,m < 20). Each of the following n lines contains m values which represent the image.

Output

For each test case, print one line which contains the number of the test case, and the recognition result: ‘Zero’ or ‘Eight’. See the samples and follow the output format.

Sample Input

Input

3
8 10
..........
..*****...
..*...*...
..*...*...
..*****...
..*...*...
..*****...
..........
6 10
..........
..*****...
..*...*...
..*...*...
..*****...
..........
10 7
.......
.......
.......
.......
..****.
..*..*.
..****.
..*..*.
..****.
.......

Output

Case 1: Eight
Case 2: Zero
Case 3: Eight



看上去很高端的题 其实很水

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <vector>
#include <iomanip>
#include <math.h>
#include <map>
using namespace std;
#define FIN     freopen("input.txt","r",stdin);
#define FOUT    freopen("output.txt","w",stdout);
#define INF     0x3f3f3f3f
#define lson    l,m,rt<<1
#define rson    m+1,r,rt<<1|1
typedef long long LL;

const int MAXN=1000+5;
char Map[25][25];
int flag;

int llss(int x,int y){
    if(Map[x][y+1]=='*')  {flag++;}
    if(Map[x][y]=='.')  return 0;
    llss(x+1,y);
}


int main()
{
    //FIN
    int T;
    scanf("%d",&T);
    for(int z=1;z<=T;z++){
        int n,m;
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n;i++)
            for(int j=1;j<=m;j++){
                //scanf("%c",Map[i][j]);
                cin>>Map[i][j];
            }

        flag=0;
        int k=0;
        for(int i=1;i<=n;i++){
            for(int j=1;j<=m;j++){
                if(Map[i][j]=='*'){
                    llss(i,j);
                    k=1;
                    break;
                }
            }
            if(k==1)  break;
        }

        if(flag==2)  printf("Case %d: Zero
",z);
        if(flag==3)  printf("Case %d: Eight
",z);
    }
}