带限制的广搜 codeforces

You are playing some computer game. One of its levels puts you in a maze consisting of n lines, each of which contains m cells. Each cell either is free or is occupied by an obstacle. The starting cell is in the row r and column c. In one step you can move one square up, left, down or right, if the target cell is not occupied by an obstacle. You can't move beyond the boundaries of the labyrinth.

Unfortunately, your keyboard is about to break, so you can move left no more than x times and move right no more than y times. There are no restrictions on the number of moves up and down since the keys used to move up and down are in perfect condition.

Now you would like to determine for each cell whether there exists a sequence of moves that will put you from the starting cell to this particular one. How many cells of the board have this property?

Input

The first line contains two integers n, m (1 ≤ n, m ≤ 2000) — the number of rows and the number columns in the labyrinth respectively.

The second line contains two integers r, c (1 ≤ r ≤ n, 1 ≤ c ≤ m) — index of the row and index of the column that define the starting cell.

The third line contains two integers x, y (0 ≤ x, y ≤ 10⁹) — the maximum allowed number of movements to the left and to the right respectively.

The next n lines describe the labyrinth. Each of them has length of m and consists only of symbols '.' and '*'. The j-th character of the i-th line corresponds to the cell of labyrinth at row i and column j. Symbol '.' denotes the free cell, while symbol '*' denotes the cell with an obstacle.

It is guaranteed, that the starting cell contains no obstacles.

Output

Print exactly one integer — the number of cells in the labyrinth, which are reachable from starting cell, including the starting cell itself.

Examples

Input

Copy

4 5
3 2
1 2
.....
.***.
...**
*....

Output

Copy

10

Input

Copy

4 4
2 2
0 1
....
..*.
....
....

Output

Copy

7

Note

Cells, reachable in the corresponding example, are marked with '+'.

First example:

+++..
+***.
+++**
*+++.

Second example:

.++.
.+*.
.++.
.++.

题意 ： 给你一个n*m 的矩阵，再给你一个起点，你可以往上下左右四个方向走，但同时又有一个往左右的步数限制，询问所有可能到达的点数有多少个？
思路分析 :
　　比赛场上直接写了一个 bfs ，pp 了，自信以为A了，今晚又可以涨分，结果....
　　其实有一点贪心的想法，就是可以上下走的时候一定要先上下走，最后再左右走，用一个双端队列即可
代码示例：

int n, m;
int sx, sy;
int sl, sr;

char mp[2005][2005];
struct node
{
    int x, y;
    int l, r;
};
deque<node>que;
bool vis[2005][2005];

int dir[4][2] = {1, 0, -1, 0, 0, 1, 0, -1};

void bfs(){
    que.push_front({sx, sy, sl, sr});
    int ans = 0;
    //vis[sx][sy] = 1;

    while(!que.empty()){
        node v = que.front(); que.pop_front();
        if (vis[v.x][v.y]) continue;
        vis[v.x][v.y] = true;
        ans++; 
        for(int i = 0; i < 4; i++){
            int fx = v.x+dir[i][0];
            int fy = v.y+dir[i][1];
            if (fx < 1 || fx > n || fy < 1 || fy > m || mp[fx][fy] == '*'||vis[fx][fy]) continue;
            
            if (i == 2) {
                if (v.r > 0) que.push_back({fx, fy, v.l, v.r-1});
            } 
            else if (i == 3) {
                if (v.l > 0) que.push_back({fx, fy, v.l-1, v.r});
            } 
            else que.push_front({fx, fy, v.l, v.r});
        }    
    }  
    printf("%d
", ans);  
}

int main() {
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    cin >> n >> m;
    cin >> sx >> sy;
    cin >> sl >> sr;
    for(int i = 1; i <= n; i++){
        scanf("%s", mp[i]+1);
    }
    bfs();
    for(int i = 1; i <= n; i++){
        for(int j = 1; j <= m; j++){
            if(mp[i][j] == '*') printf("*");
            else if (vis[i][j]) printf("+");
            else printf(".");
        }
        printf("
");
    }
    return 0;
}

 2 . 用类似最短路的写法，每次左右走的时候步数+1，上下走的时候步数不变

代码示例：

const int inf = 0x3f3f3f3f;

int n, m;
int sx, sy, sl, sr;
char mp[2005][2005];
struct node
{
    int x, y;
    int l, r;
    int sum;
    bool operator< (const node &v)const{
        return sum < v.sum;
    }
};
priority_queue<node>que;
bool vis[2005][2005];
int bu[2005][2005];
int dir[4][2] = {1, 0, -1, 0, 0, 1, 0, -1};

void bfs() {
    que.push({sx, sy, sl, sr, sl+sr});
    memset(vis, false, sizeof(vis));
    memset(bu, inf, sizeof(bu));
    int ans = 0;
    bu[sx][sy] = 0;
    
    while(!que.empty()){
        node v = que.top(); que.pop();
        
        if (!vis[v.x][v.y]) ans++;
        vis[v.x][v.y] = 1;
        for(int i = 0; i < 4; i++){
            int fx = dir[i][0]+v.x;
            int fy = dir[i][1]+v.y;
    
            if (fx < 1 || fx > n || fy < 1 || fy > m || mp[fx][fy] == '*') continue;
            if (bu[v.x][v.y] >= bu[fx][fy]) continue;
                bu[fx][fy] = bu[v.x][v.y];
            
            if ((i == 2) || (i == 3)) bu[fx][fy]++;
            if (i == 2) {
                if (v.r) que.push({fx, fy, v.l, v.r-1, v.l+v.r-1});
            }
            else if (i == 3) {
                if (v.l) que.push({fx, fy, v.l-1, v.r, v.l+v.r-1});
            }
            else que.push({fx, fy, v.l, v.r, v.l+v.r});  
        }
    }    
    printf("%d
", ans);
}

int main() { 
    cin >> n >> m;
    cin >> sx >> sy;
    cin >> sl >> sr;
    for(int i = 1; i <= n; i++) scanf("%s", mp[i]+1);
    bfs();
    return 0;
}