Usaco Training [2.1] The Castle 搜索

传送门

题目的输出的4个信息

前两个很容易，dfs，bfs都可以，图怎么建都可以

后两个在搜索的时候记录belong[i][j]和已有的size即可

代码应该比不少题解清晰吧

 #include <bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 #define rep(i, a, b) for (int i = a; i <= b; ++i)

 const int N = ;

 int m, n, a[N][N], bel[N][N], siz[N * N], cnt, mx;

 int b[N][N][], dir[][] = {{, -}, {-, }, {, }, {, }};

 bool vis[N][N];

 void dfs(int x, int y) {
     if (!vis[x][y] && x >  && x <= m && y >  && y <= n) {
         siz[cnt]++;
         vis[x][y] = ;
         bel[x][y] = cnt;
         rep(i, , ) if (!b[x][y][i]) {
             dfs(x + dir[i][], y + dir[i][]);
         }
     }
 }

 int main() {
     scanf("%d%d", &n, &m);

     rep(i, , m) rep(j, , n) {
         scanf("%d", &a[i][j]);
         int t = a[i][j];
         b[i][j][] = (t %  > ); t -= t % ;
         b[i][j][] = (t %  > ); t -= t % ;
         b[i][j][] = (t %  > ); t -= t % ;
         b[i][j][] = (t %  > );
     }

     rep(i, , m) rep(j, , n) if (!vis[i][j]) {
         ++cnt;
         dfs(i, j);
         mx = max(mx, siz[cnt]);
     }

     printf("%d\n%d\n", cnt, mx);

     int dx, dy; char ch;
     rep(j, , n) for(int i = m; i >= ; --i)  rep(k, , ) {
         if (bel[i][j] != bel[i + dir[k][]][j + dir[k][]]) {
             if (siz[bel[i][j]] + siz[bel[i + dir[k][]][j + dir[k][]]] > mx) {
                 mx = siz[bel[i][j]] + siz[bel[i + dir[k][]][j + dir[k][]]];
                 if (k == ) dx = i + dir[k][], dy = j + dir[k][], ch = 'E';
                 else if (k == ) dx = i, dy = j, ch = 'N';
                 else if (k == ) dx = i, dy = j, ch = 'E';
                 else dx = i + dir[k][], dy = j + dir[k][], ch = 'N';
             }
         }
     }
     printf("%d\n%d %d %c\n", mx, dx, dy, ch);
     return ;
 }