Codeforces 598D Igor In the Museum 离线处理+bfs

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本文介绍了一个有趣的问题：如何在一个由空地和障碍组成的矩形博物馆中，从指定起点出发尽可能多地观看墙上的画作。文章提供了算法思路及实现代码，采用离线处理的方式避免重复搜索，确保效率。

Description

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.

Sample Input

Input

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

Output

6
4
10

Input

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

Output

题意：给你一副n*m的图，k次询问，保证每次询问都在“.”，问每次询问能看到多少画？

思路：一看到题目，不是个简单的bfs吗？扩展4个方向如果碰到“*”就+1，但是并不能k次询问，k次bfs。因此TLE了好几次。

所以要先遍历整个图，找每个“.”的位置都能看到的画，所以只要将连通的点存储起来，最后一并赋值，离线处理一下。k次询问直接输出即可。

因为粗心，将v[MAX]，val[MAX]数组一并写成char型的，导致wa了几次。(哭.....)

附上AC代码：

#include<stdio.h>
#include<string.h>
#include<queue>
using namespace std;
const int MAX=1005;
char Map[MAX][MAX];
int v[MAX][MAX],val[MAX][MAX];
int chan[4][2]= {1,0,-1,0,0,1,0,-1},sum,n,m,anc;
struct node
{
    int x,y;
} D[1000000];
void bfs(int x,int y)
{
    sum=0;
    anc=1;
    node now,now1,next;
    now1.x=x,now1.y=y;
    queue<node> Q;
    Q.push(now1);
    v[x][y]=1;
    D[0].x=x,D[0].y=y;
    while(!Q.empty())
    {
        now=Q.front();
        //printf("%d %d\n",now.x,now.y);
        Q.pop();
        int xx,yy;
        for(int i=0; i<4; i++)
        {
            xx=now.x+chan[i][0];
            yy=now.y+chan[i][1];
            if(xx>=0&&xx<n&&yy>=0&&yy<m&&!v[xx][yy]&&Map[xx][yy]!='*')
            {
                next.x=xx;
                next.y=yy;
                Q.push(next);
                D[anc].x=xx;
                D[anc++].y=yy;
                v[xx][yy]=1;
            }
            if(xx>=0&&xx<n&&yy>=0&&yy<m&&!v[xx][yy]&&Map[xx][yy]=='*')
            {
                sum++;
            }
        }
    }
    return ;
}
int main()
{
    int k,x,y;
    while(~scanf("%d%d%d",&n,&m,&k))
    {
        memset(v,0,sizeof(v));
        memset(val,0,sizeof(val));
        for(int i=0; i<n; i++)
            scanf("%s",Map[i]);
        for(int i=0; i<n; i++)
        {
            for(int j=0; j<m; j++)
            {
                if(!v[i][j]&&Map[i][j]=='.')
                {
                    bfs(i,j);
                    //printf("%d  %d  %d\n",sum,i,j);
                    for(int i=0; i<anc; i++)
                    {
                        val[D[i].x][D[i].y]=sum;
                    }
                }
            }
        }
        for(int i=0; i<k; i++)
        {
            scanf("%d%d",&x,&y);
            //printf("%d\n",v[x-1][y-1]);
            printf("%d\n",val[x-1][y-1]);
        }
    }
}