[【USACO】The Castle（dfs+枚举）

思路很好像，卡了我很久的就是当最大房间一样的时候判断输出哪个的条件， = = 简直无情

/*
ID: 18906421
LANG: C++
PROG: castle
*/
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn = 55;
int mat[maxn][maxn][2] = {0}; // 0 1  下  右
int n,m,vis[maxn][maxn];
int max1 = 0,max2 = 0,cnt = 0;
void debug(){
    for(int i = 0; i < n; i++){
        for(int j = 0; j < m; j++)
            printf("[%d,%d],(%d %d)",i,j,mat[i][j][0],mat[i][j][1]);
        puts("");
    }
}
void change(int x,int  y,int v){
    if(v == 8 || v == 9 || v == 10 || v == 12 || v == 11 || v == 13 || v == 14 || v == 15)
        mat[x][y][0] = 1;
    if(v == 4 || v == 5 || v == 6  || v == 12 || v == 7  || v == 14 || v == 13 || v == 15)
        mat[x][y][1] = 1;
}
int dfs(int x,int y){
    vis[x][y] = 1;
    int ans = 1;
    if(!mat[x][y][0] && x + 1 <  n && !vis[x + 1][y]) ans += dfs(x + 1,y);      // 下
    if(!mat[x][y][1] && y + 1 <  m && !vis[x][y + 1]) ans += dfs(x,y + 1);      // 右
    if(x - 1 >= 0 && !mat[x - 1][y][0] && !vis[x - 1][y]) ans += dfs(x - 1,y);  // 上
    if(y - 1 >= 0 && !mat[x][y - 1][1] && !vis[x][y - 1]) ans += dfs(x,y - 1);  // 左
    return ans;
}
void solve(int &max_size,int &cc){
    memset(vis,0,sizeof(vis));
    for(int i = 0; i < n; i++)
        for(int j = 0; j < m; j++)
            if(!vis[i][j]){
                max_size = max(max_size,dfs(i,j));
                cc ++;
            }
}
int main(){
    freopen("castle.in","r",stdin);
    freopen("castle.out","w",stdout);
    scanf("%d%d",&m,&n);
    for(int i = 0; i < n; i++)
        for(int j = 0; j < m; j++){
            int x;
            scanf("%d",&x);
            change(i,j,x);
        }
    //debug();
    solve(max1,cnt); //获得最大房间
    printf("%d\n%d\n",cnt,max1);
    int dirr,pos_x,pos_y,max_size = max1,cc;
    for(int j = m - 1; j >= 0; j--)
        for(int i = 0; i < n; i++){
            for(int k = 1; k >= 0; k --)
                if(mat[i][j][k]){       //k 为 0 破坏下面 1 破坏右边
                    max2 = 0;cc = 0;
                    mat[i][j][k] = 0;
                    solve(max2,cc);
                    mat[i][j][k] = 1;
                    int xx = k == 0 ? i + 1 : i;
                    int yy = j;
                    if(max2 > max_size){
                        pos_x = xx; pos_y = yy;
                        max_size = max2;
                        dirr = k;
                    }
                    else if(max2 == max_size){
                        if(yy < pos_y || (yy == pos_y && xx > pos_x)){
                            pos_x = xx; pos_y = yy;
                            max_size = max2;
                            dirr = k;
                        }
                    }
                }
        }
    if(dirr == 1)
        printf("%d\n%d %d %c\n",max_size,pos_x + 1,pos_y + 1,'E');
    else
        printf("%d\n%d %d %c\n",max_size,pos_x + 1,pos_y + 1,'N');
    return 0;
}