cf-Igor In the Museum (dfs)

博物馆寻宝：最优路径与视觉艺术的探索

最新推荐文章于 2020-07-08 19:18:04 发布

转载最新推荐文章于 2020-07-08 19:18:04 发布 · 44 阅读

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原文链接：http://www.cnblogs.com/soTired/p/5247806.html

本文深入探讨了Igor在博物馆中寻找尽可能多的艺术品的策略，通过构建模型和算法来优化他的参观路径。文章详细介绍了博物馆布局、艺术品分布以及Igor的移动限制，并提供了一种计算最优路径的方法，以最大化艺术品的观看数量。

Igor is in the museum and he wants to see as many pictures as possible.

Museum can be represented as a rectangular field of n × m cells. Each cell is either empty or impassable. Empty cells are marked with '.', impassable cells are marked with '*'. Every two adjacent cells of different types (one empty and one impassable) are divided by a wall containing one picture.

At the beginning Igor is in some empty cell. At every moment he can move to any empty cell that share a side with the current one.

For several starting positions you should calculate the maximum number of pictures that Igor can see. Igor is able to see the picture only if he is in the cell adjacent to the wall with this picture. Igor have a lot of time, so he will examine every picture he can see.

Input

First line of the input contains three integers n, m and k(3 ≤ n, m ≤ 1000, 1 ≤ k ≤ min(n·m, 100 000)) — the museum dimensions and the number of starting positions to process.

Each of the next n lines contains m symbols '.', '*' — the description of the museum. It is guaranteed that all border cells are impassable, so Igor can't go out from the museum.

Each of the last k lines contains two integers x and y (1 ≤ x ≤ n, 1 ≤ y ≤ m) — the row and the column of one of Igor's starting positions respectively. Rows are numbered from top to bottom, columns — from left to right. It is guaranteed that all starting positions are empty cells.

Output

Print k integers — the maximum number of pictures, that Igor can see if he starts in corresponding position.

Examples

input

5 6 3
******
*..*.*
******
*....*
******
2 2
2 5
4 3

output

6
4
10

input

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

output

8

好吧， 我没读懂题；

#include <cstdio>
#include <cstring>
#define M 1001
int id, ans;
int n, m;
char G[M][M]; int v[M][M], r[M*M]; 
int ac[4][2]={0, 1, 0, -1, -1, 0, 1, 0};
void dfs(int x, int y)
{
    v[x][y]=id;                 //标记数组； 
    for(int i=0; i<4; i++)
    {
        int nx=x+ac[i][0];
        int ny=y+ac[i][1];
        if(nx>0&&nx<=n&&ny>0&&ny<=m&&!v[nx][ny])
        {
            if(G[nx][ny]=='*')
                ans++;
            else
                dfs(nx, ny);
        }
    }
}
int main()
{
    int k;
    while(scanf("%d%d%d", &n, &m, &k)!=EOF)
    {
        for(int i=1; i<=n; i++)
            scanf("%s", G[i]+1);
        memset(v, 0, sizeof(v));
        id=0;
        for(int i=1; i<=n; i++)
        {
            for(int j=1; j<=m; j++)
            {
                if(G[i][j]=='.'&&!v[i][j])
                {
                    ans=0;
                    id++;
                    dfs(i, j);
                    r[id]=ans;    
                //    printf("%d\n", ans);                
                }
            }
        }
        while(k--)
        {
            int x, y;
            scanf("%d%d", &x, &y);
            printf("%d\n", r[v[x][y]]);
        }
    }
    return 0;
}