POJ 3620:Avoid The Lakes(DFS)

可以理解为判断连通块的个数~~DFS解决

#include"cstdio"
#include"iostream"
#include"cstring"
#define MAXN 105
using namespace std;
int mat[MAXN][MAXN],vis[MAXN][MAXN];
int dfs(int x,int y)
{   int ans=0;
    if(!mat[x][y]||vis[x][y]) return 0;
    ans++;vis[x][y]=1;
    ans+=dfs(x+1,y);ans+=dfs(x-1,y);ans+=dfs(x,y+1);ans+=dfs(x,y-1);
    return ans;
}
int main()
{   int n,m,k;
    while(scanf("%d%d%d",&n,&m,&k)!=EOF)
    {   memset(mat,0,sizeof(mat));
        memset(vis,0,sizeof(vis));
        for(int i=0;i<k;i++)
        {   int a,b;
            scanf("%d%d",&a,&b);
            mat[a][b]=1;
        }
        int ans=0;
        for(int i=1;i<=n;i++)
            for(int j=1;j<=m;j++)
               ans=max(dfs(i,j),ans);
        printf("%d
",ans);
    }
    return 0;
}