Best Grass

Description

Bessie is planning her day of munching tender spring grass and is gazing
out upon the pasture which Farmer John has so lovingly partitioned into a
grid with R (1 <= R <= 100) rows and C (1 <= C <= 100) columns. She wishes
to count the number of grass clumps in the pasture.

Each grass clump is shown on a map as either a single '#' symbol or perhaps
two '#' symbols side-by-side (but not on a diagonal). Given a map of the
pasture, tell Bessie how many grass clumps there are.

By way of example, consider this pasture map where R=5 and C=6:

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

This pasture has a total of 5 clumps: one on the first row, one that spans
the second and third row in column 2, one by itself on the third row, one
that spans columns 4 and 5 in row 4, and one more in row 5.

Input

* Line 1: Two space-separated integers: R and C

* Lines 2..R+1: Line i+1 describes row i of the field with C
        characters, each of which is a '#' or a '.'

Output

* Line 1: A single integer that is the number of grass clumps Bessie
        can munch

• Sample Input
• Raw

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

• Sample Output
• Raw

5

#include<cstdio>
#include<iostream>
using namespace std;
int x[4]={1,0,-1,0},y[4]={0,1,0,-1};//四个方向
char Map[105][105];
int C,R,ans;
void DFS(int i,int j){
    Map[i][j]='.';
    for(int t=0;t<4;t++){
        int xn=i+x[t];
        int yn=j+y[t];
        if(xn>=0&&xn<R&&yn>=0&&yn<C&&Map[xn][yn]=='#')
            DFS(xn,yn);
    }
}
int main ()
{
    while(~scanf("%d%d",&R,&C)){
        ans=0;
        for(int i=0;i<R;i++)
            for(int j=0;j<C;j++)
                cin>>Map[i][j];
        for(int i=0;i<R;i++)
            for(int j=0;j<C;j++){
                if(Map[i][j]=='#'){
                    DFS(i,j);
                    ans++;
                }
            }
        printf("%d
",ans);
    }
    return 0;
}