CodeForces 327D(DFS)

淩亂°似流年 2022-06-06 05:44 133阅读 0赞

问题描述:

After too much playing on paper, Iahub has switched to computer games. The game he plays is called "Block Towers". It is played in a rectangular grid with n rows and mcolumns (it contains n × m cells). The goal of the game is to build your own city. Some cells in the grid are big holes, where Iahub can't build any building. The rest of cells are empty. In some empty cell Iahub can build exactly one tower of two following types:

1.  Blue towers. Each has population limit equal to 100.
2.  Red towers. Each has population limit equal to 200. However, it can be built in some cell only if in that moment at least one of the neighbouring cells has a Blue Tower. Two cells are neighbours is they share a side.

Iahub is also allowed to destroy a building from any cell. He can do this operation as much as he wants. After destroying a building, the other buildings are not influenced, and the destroyed cell becomes empty (so Iahub can build a tower in this cell if needed, see the second example for such a case).

Iahub can convince as many population as he wants to come into his city. So he needs to configure his city to allow maximum population possible. Therefore he should find a sequence of operations that builds the city in an optimal way, so that total population limit is as large as possible.

He says he's the best at this game, but he doesn't have the optimal solution. Write a program that calculates the optimal one, to show him that he's not as good as he thinks.

Input

The first line of the input contains two integers n and m (1 ≤ n, m ≤ 500). Each of the next n lines contains m characters, describing the grid. The j\-th character in the i\-th line is '.' if you're allowed to build at the cell with coordinates (i, j)a tower (empty cell) or '\#' if there is a big hole there.

Output

Print an integer k in the first line (0 ≤ k ≤ 106) — the number of operations Iahub should perform to obtain optimal result.

Each of the following k lines must contain a single operation in the following format:

1.  «B x y» (1 ≤ x ≤ n, 1 ≤ y ≤ m) — building a blue tower at the cell (x, y);
2.  «R x y» (1 ≤ x ≤ n, 1 ≤ y ≤ m) — building a red tower at the cell (x, y);
3.  «D x y» (1 ≤ x ≤ n, 1 ≤ y ≤ m) — destroying a tower at the cell (x, y).

If there are multiple solutions you can output any of them. Note, that you shouldn't minimize the number of operations.

Example

Input

2 3
    ..#
    .#.

Output

4
    B 1 1
    R 1 2
    R 2 1
    B 2 3

Input

1 3
    ...

Output

5
    B 1 1
    B 1 2
    R 1 3
    D 1 2
    R 1 2

题目题意:在所给的图上建立出价值最大的塔，绿塔可以建立在任意'.'上（价值100），红塔必须建在'.'上且建的时候旁边必须要有绿塔，建好了没有要求（价值200），可以拆毁任何塔！

题目分析:因为红塔的价值比绿塔价值高，所以要尽可能建红塔。但是它的要求比较高，建的时候必须与绿塔相连。我们首先把所有的点都建绿塔，然后遍历图，找到一个'.'然后就去dfs(搜索它旁边是否有绿塔).其实画一个图我们就可以明白一个由n(n>1)个绿塔构成的联通块，肯定可以建立n-1座红塔，只有最后一个绿塔拆了不能建成红塔，因为没有绿塔相连。

代码如下:

#include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<queue>
    #include<stack>
    using namespace std;
    
    const int maxn=500+10;
    struct note
    {
        int x,y;
    };
    queue<struct note> res;
    stack<struct note> ans;
    int n,m,step;
    char mmap[maxn][maxn];
    int Next[4][2]={
       {-1,0},{1,0},{0,-1},{0,1}};
    
    void dfs(int x,int y)
    {
        if (step) step=0;
        else {
            struct note cur;
            cur.x=x;cur.y=y;
            ans.push(cur);
        }
        mmap[x][y]='#';
        for (int i=0;i<4;i++) {
            int dx=x+Next[i][0];
            int dy=y+Next[i][1];
            if (dx>=1&&dx<=n&&dy>=1&&dy<=m&&mmap[dx][dy]=='.')
                dfs(dx,dy);
        }
    }
    int main()
    {
        while (scanf("%d%d",&n,&m)!=EOF) {
            for (int i=1;i<=n;i++) {
                for (int j=1;j<=m;j++) {
                    cin>>mmap[i][j];
                    if (mmap[i][j]=='.') {
                      struct note cur;
                      cur.x=i;cur.y=j;
                      res.push(cur);//保存点 建成绿塔
                    }
                }
            }
            for (int i=n;i>=1;i--) {
                for (int j=m;j>=1;j--) {
                    if (mmap[i][j]=='.') {
                        step=1;//把第一个保留，其余建成红塔
                        dfs(i,j);
                    }
                }
            }
            step=res.size()+ans.size()*2;
            printf("%d\n",step);
            while (res.size()) {
                struct note cur=res.front();
                res.pop();
                printf("B %d %d\n",cur.x,cur.y);
            }
            while (ans.size()) {
                struct note cur=ans.top();
                ans.pop();
                printf("D %d %d\nR %d %d\n",cur.x,cur.y,cur.x,cur.y);
            }
        }
        return 0;
    }