17: The best grass, the best grass

17: The best grass, the best grass
17: The best grass

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Total time limit:

10000 ms

 

Time limit for a single test point:

1000 ms

 

Memory limit:

65536kB

Description

Bessie, the cow, plans to enjoy the soft spring grass. The grass is distributed in the farm of column C in the R row. It wants to calculate the number of grass in the pasture.

In a farm map, each grass is either a single "#" or two adjacent "#" with a public edge. Calculate the number of grass trees in a given farm map.

For example, consider the farm map with 5 rows and 6 Columns

.#....
..#...
..#..#
...##.
.#....

This farm has five grass trees: one in the first row, one in the second column across the second and third rows, one in the third row, and the other in the fourth row across the fourth and fifth columns, the last one is on the fifth line.

 

Input

The first line contains two integers, R and C, separated by a single space.
Next, the R Line describes the farm map with C characters in each line. There are only two types of characters: "#" and. (1 <= R, C <= 100)

Output

Output an integer to indicate the number of grass.

Sample Input

5 6.#......#.....#..#...##..#....

Sample output

5

Source

USACO Open 2008 Bronze

 

 

 1 #include<iostream> 2 #include<cstdio> 3 #include<cstdlib> 4 using namespace std; 5 char a[1001][1001]; 6 int now=1; 7 int m_tot=0; 8 int z_tot=0; 9 int ans=0;10 int main() 11 {12     int n,m;13     cin>>n>>m;14     for(int i=0;i<n;i++)15     {16         for(int j=0;j<m;j++)17         {18             cin>>a[i][j];19         }20     }21     for(int i=0;i<n;i++)22     {23         for(int j=0;j<m;j++)24         {25             if(a[i][j]=='#')26             {27                 if(a[i+1][j]=='#')28                 {29                     a[i][j]='.';30                     a[i+1][j]='.';31                     ans++;32                 }33                 else if(a[i-1][j]=='#')34                 {35                     a[i][j]='.';36                     a[i+1][j]='.';37                     ans++;38                 }39                 else if(a[i][j+1]=='#')40                 {41                     a[i][j]='.';42                     a[i][j+1]='.';43                     ans++;44                 }45                 else if(a[i][j-1]=='#')46                 {47                     a[i][j]='.';48                     a[i+1][j]='.';49                     ans++;50                 }51                 else52                 {53                     a[i][j]='.';54                     ans++;55                 }56             }57         }58     }59     cout<<ans;60     return 0;61 }