CF 327D - 走看看

CF 327D

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 m columns (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:

Blue towers. Each has population limit equal to 100.
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:

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

Sample test(s)
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

中文意思：

就是在空的地方(字符'-')建造房子，蓝色房子是100人口，红色房子是200人口（建造红色房子时，旁边必须有蓝色房子），要求人口最大，只需输出可用的解就可以了。k是可行解的步数。

题解：

此题就是将尽可能的建造红色房子，但要求在建造红色房子时，它旁边必须有个蓝色房子。但输出无顺序要求。首先我们将图分成几块（被#完全隔开）。你可以发现这样一点。你将所有的先全部建造成蓝色，然后再慢慢的从边际（也就是叶子）开始逐个的将此变成红色。看样例3，你就明白了。如此，每块分割开的一个个块将最后变成一个蓝的房子跟红色的其他房子。我们很容易知道，这时居住人口是最多的。因此，DFS即可解决这个问题。

因为n<=m<=500。我们的最大步数是3*m*n<k<=10^6次。所以这样的解是可以的。

我们的k值就是空房数量*3-蓝色房子（一个块只有1个）*2；k可以直接在DFS时出来（先压点到栈）。

之后不断弹出栈中元素，分别输出 D操作跟R操作即可。。

/*
 * @author ipqhjjybj
 * @date  20130705
 *
 */
#include <cstdio>
#include <cstring>
#include <iostream>
#include <stack>
#define clr(x,k) memset((x),(k),sizeof(x))
#define MAX(a,b) ((a)>(b)?(a):(b))
using namespace std;

char s[505][505];
int n,m;
struct node{
    int x,y;
    bool f;   //表示是否为一个块的开头
    node(){};
    node(int xx,int yy,bool first){
        x=xx,y=yy,f=first;
    };
};
stack<node> sn;
bool visit[505][505];
int k;
int x1[4]={0,0,1,-1};
int y1[4]={1,-1,0,0};
#define legal(x,a) (1<=x&&x<=a)
void dfs(int xx,int yy){
    int _x,_y,i;
    for(i=0;i<4;i++){
        _x=xx+x1[i],_y=yy+y1[i];
        if(legal(_x,n)&&legal(_y,m)&&!visit[_x][_y]&&s[_x][_y]=='.'){
            visit[_x][_y]=true;
            sn.push(node(_x,_y,false));
            dfs(_x,_y);
        }
    }
}
int main(){
   // freopen("D.in","r",stdin);
    int i,j;
    while(scanf("%d %d",&n,&m)!=EOF){
        getchar();
        for(i=1;i<=n;i++)
            gets(s[i]+1);
        sn.empty();
        k=0;
        clr(visit,false);
        for(i=1;i<=n;i++)
            for(j=1;j<=m;j++){
                if(!visit[i][j]&&s[i][j]=='.'){
                    k-=2;
                    visit[i][j]=true;
                    sn.push(node(i,j,true));
                    dfs(i,j);
                }
            }
        k+=sn.size()*3;
        printf("%d
",k);
        for(i=1;i<=n;i++)
            for(j=1;j<=m;j++)
                if(visit[i][j])
                    printf("B %d %d
",i,j);
        node t;
        while(!sn.empty()){
            t = sn.top(); sn.pop();
            if(!t.f){
                printf("D %d %d
",t.x,t.y);
                printf("R %d %d
",t.x,t.y);
            }
        }
    }
    return 0;
}