18: Tumor area, 18 tumor Area

18: Tumor area, 18 tumor Area
18: Tumor Area

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Total time limit:

1000 ms

 

Memory limit:

65536kB

Description

On a square grayscale image, the tumor is a rectangular area, and the pixel of the tumor edge is represented by 0 in the image. Other tumors and vertices outside the tumor are expressed as 255. Now you need to write a program to calculate the number of pixels inside the tumor (excluding vertices on the tumor edge ). It is known that the tumor edge is parallel to the image edge.

Input

There is only one test sample. The first line has an integer n, indicating the side length of the square image. Each row in the next n Rows has n integers. The values are 0 or 255. Integers are separated by a space. It is known that n is not greater than 1000.

Output

Output a row containing an integer, which is the number of pixels in the required tumor.

Sample Input

5255 255 255 255 255255 0 0 0 255255 0 255 0 255255 0 0 0 255255 255 255 255 255

Sample output

1

Prompt

If you use a static array to represent image data, you need to define the array as a global variable.

Source

2005 ~ 2006 Final Examination of Introduction to computing in the Medical Department

 1 #include<iostream> 2 using namespace std; 3 int a[1001][1001]; 4 int now=1; 5 int tot=0; 6 int hang,lie; 7 int l_h; 8 int l_l; 9 int ans=0;10 int main() 11 {12     int n;13     cin>>n;14     for(int i=1;i<=n;i++)15     {16         for(int j=1;j<=n;j++)17         {18             cin>>a[i][j];19         }20     }21     for(int i=1;i<=n;i++)22     {23         for(int j=1;j<=n;j++)24         {25             if(a[i-1][j]==0&&a[i-1][j-1]==0&&a[i][j-1]==0&&i!=1&&j!=1)26             {27                 hang=i;28                 lie=j;29             }30         }31     }32     for(int i=hang;i<=9999;i++)33     {34         for(int j=lie;j<=9999;j++)35         {36             if(a[i][j]!=0)37             {38                 l_l++;39             }40             else break;41         }42         break;43     }44     for(int i=hang;i<=9999;i++)45     {46         if(a[i][lie]!=0)47         {48             l_h++;49         }50         else break;51     }52     cout<<l_h*l_l;53     return 0;54 }