DFS PKU 1562

标签：dfs

简单的DFS

Description

Input

Output

Sample Input

1 1
*
3 5
*@*@*
**@**
*@*@*
1 8
@@****@*
5 5 
****@
*@@*@
*@**@
@@@*@
@@**@
0 0

Sample Output

0
1
2
2

#include <stdio.h>
#include <stdlib.h>
#include <malloc.h>
#include <limits.h>
#include <ctype.h>
#include <string.h>
#include <string>
#include <math.h>
#include <algorithm>
#include <iostream>
#include <queue>
#include <stack>
#include <deque>
#include <vector>
#include <set>
//#include <map>
using namespace std;
#define MAXN 100 + 10
char map[MAXN][MAXN];
int vis[MAXN][MAXN];
int m,n;
//int count;
int xx[8] = {-1,-1,0,1,1,1,0,-1};
int yy[8] = {0,1,1,1,0,-1,-1,-1};

void DFS(int x,int y){
    int i;
    for(i=0;i<8;i++){
        int dx = x+xx[i];
        int dy = y+yy[i];
        if(dx>=0 && dx<m && dy>=0 && dy<n){
            if(map[dx][dy]=='@' && vis[dx][dy]==0){
                vis[dx][dy] = 1;
                DFS(dx,dy);
                //vis[dx][dy] = 0;
            }
        }
    }

    return ;
}

int main(){
    int i,j;
    int ans;

    while(~scanf("%d%d",&m,&n)){
        if(m==0 && n==0){
            break;
        }
        for(i=0;i<m;i++){
            scanf("%s",map[i]);
        }
        ans = 0;
        memset(vis,0,sizeof(vis));
        for(i=0;i<m;i++){
            for(j=0;j<n;j++){
                if(map[i][j]=='@' && vis[i][j]==0){
                    vis[i][j] = 1;
                    ans++;
                    DFS(i,j);
                    //vis[i][j] = 0;
                }
            }
        }
        printf("%d\n",ans);
    }
}

DFS PKU 1562

标签：dfs

原文地址：http://blog.csdn.net/zcr_7/article/details/39716979