hdu 1198 (并查集 or dfs) Farm Irrigation

题目:http://acm.hdu.edu.cn/showproblem.php?pid=1198

有题目图11种土地块，块中的绿色线条为土地块中修好的水渠，现在一片土地由上述的各种土地块组成，需要浇水，问需要打多少口井

比如

ADC
FJK
IHE是下图这种情况

需要打三口井,其实想多了就是集合合并差不多,用并查集可以解决,将每个格子按顺序是从0到n*m-1,然后根据连接情况判断是否相连然后合并

第一次写成了暴力的,虽然一遍过了,但是觉得太长了,然后又改成了下面简洁并查集的,还有下下面的dfs的

并查集

 #include<cstdio>
 using namespace std;
 char yj[][];
 int father[*+];
 int dir[][]={{,,,},{,,,},{,,,},
                 {,,,},{,,,},{,,,},
                 {,,,},{,,,},{,,,},
                 {,,,},{,,,}};
 void give(int x)
 {
     for (int i=;i<x;i++)
         father[i]=i;
 }
 int _find(int x)
 {
     if (x==father[x]) return father[x];
     father[x]=_find(father[x]);
     return father[x];
 }
 int merge(int x,int y)
 {
     int sx=_find(x);
     int sy=_find(y);
     if (sx!=sy)
         father[sx]=sy;
 }
 int main()
 {
     int n,m,i,j;
     while (~scanf("%d %d",&n,&m))
     {
         if (n==-&&m==-) break;
         for (i=;i<n;i++)
             scanf("%s",yj[i]);
         give(n*m);
         for (i=;i<n;i++)
         {
             for (j=;j<m;j++)
             {
                 int x=yj[i][j]-'A';
                 int y=yj[i][j+]-'A';
                 if (dir[x][]==&&dir[y][]==&&j+<m)
                     merge(i*m+j,i*m+j+);
                 y=yj[i+][j]-'A';
                 if (dir[x][]==&&dir[y][]==&&i+<n)
                     merge(i*m+j,(i+)*m+j);
             }
         }
         int sum=;
         for (i=;i<n*m;i++)
            if (father[i]==i)
              sum++;
         printf("%d\n",sum);
     }
     return ;
 }

dfs

 #include<cstdio>
 #include<cstring>
 using namespace std;
 int dx[]={,-,,};//下上左右
 int dy[]={,,-,};
 char yj[][];
 int ans,vis[][],n,m;
 int dir[][]={{,,,},{,,,},{,,,},
                 {,,,},{,,,},{,,,},
                 {,,,},{,,,},{,,,},
                 {,,,},{,,,}};
 int check(int x,int y,int di)
 {
     if (dir[x][]&&dir[y][]&&di==) return ;
     if (dir[x][]&&dir[y][]&&di==) return ;
     if (dir[x][]&&dir[y][]&&di==) return ;
     if (dir[x][]&&dir[y][]&&di==) return ;
     return ;
 }
 void dfs(int x,int y)
 {
     int i;
     if (vis[x][y]) return;
     vis[x][y]=;
     //printf("%d %d\n",x,y);
     for (i=;i<;i++)
     {
         int sx=x+dx[i];
         int sy=y+dy[i];
         if (sx<||sx>=n||sy<||sy>=m) continue;
         if (vis[sx][sy]) continue;
         int q=yj[sx][sy]-'A';
         int w=yj[x][y]-'A';
         if (check(q,w,i))
             dfs(sx,sy);
     }
 }
 int main()
 {
     int i,j;
     while (~scanf("%d %d",&n,&m))
     {
         if (n==-&&m==-) break;
         for (i=;i<n;i++)
             scanf("%s",yj[i]);
         memset(vis,,sizeof(vis));
         ans=;
         for (i=;i<n;i++)
         {
             for (j=;j<m;j++)
             {
                 if (vis[i][j]) continue;
                 dfs(i,j);
                 ans++;
             }
         }
         printf("%d\n",ans);
     }
     return ;
 }