POJ 1979 Red and Black DFS

简单DFS

注意边界就行了。

// #pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<iostream>
#include<sstream>
#include<cmath>
#include<climits>
#include<string>
#include<map>
#include<queue>
#include<vector>
#include<stack>
#include<set>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
#define pb(a) push_back(a)
#define INF 0x1f1f1f1f
#define lson idx<<1,l,mid
#define rson idx<<1|1,mid+1,r

void debug()
{
#ifdef ONLINE_JUDGE
#else
    freopen("d:\in.txt","r",stdin);
    freopen("d:\out1.txt","w",stdout);
#endif
}
char getch()
{
    char ch;
    while((ch=getchar())!=EOF)
    {
        if(ch!=' '&&ch!='
')return ch;
    }
    return EOF;
}
char da[25][25];
int n,m;
int num;
int dx[]={-1,0,1,0};
int dy[]={0,1,0,-1};
int vis[25][25];
int dfs(int x,int y)
{
    vis[x][y]=1;
    num++;
    for(int d=0;d<4;d++)
    {
        int nx=x+dx[d],ny=y+dy[d];
        if(da[nx][ny]=='.'&&!vis[nx][ny]&&nx>=1&&nx<=n&&ny>=1&&ny<=m)
            dfs(nx,ny);
    }
    return 0;
}
int main()
{
    while(scanf("%d%d",&m,&n)!=EOF&&(n||m))
    {
        for(int i=1; i<=n; i++)
            scanf("%s",da[i]+1);
        num=0;
        int bx,by;
        for(int i=1; i<=n; i++)
            for(int j=1; j<=m; j++)
            {
                if(da[i][j]=='@')
                {
                    bx=i;
                    by=j;
                    break;
                }
            }
        memset(vis,0,sizeof(vis));
        dfs(bx,by);
        printf("%d
",num);
    }
    return 0;
}
View Code
原文地址:https://www.cnblogs.com/BMan/p/3238104.html