CodeForces 598D Igor In the Museum【dfs】

最新推荐文章于 2019-08-16 08:54:00 发布

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最新推荐文章于 2019-08-16 08:54:00 发布

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分类专栏： CodeForces 深度优先搜索文章标签： dfs codeforces

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本文介绍了一个基于博物馆探索的游戏问题，玩家需要在一个由空地和障碍组成的网格地图中移动，目标是尽可能多地观察到墙上的画作。文章详细解释了游戏规则，并提供了一段使用深度优先搜索(DFS)算法来解决该问题的C++代码实现。

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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:给你一幅图,k次问你'.'的坐标下'.'连通块周围有多少个'*'
    类型:dfs
    分析:用TimeClock来区分不同连通块,对当前的TimeClock,把dfs到的所有位置(x,y)标记为time[x][y]=TimeClock,
         dfs结束后,用数组num[dfsclock]记录连通块周围*的个数.
*/

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
const int maxn = 1005;
char Map[maxn][maxn];
int Time[maxn][maxn];
int vis[maxn][maxn];
int dx[4]={0,0,1,-1},dy[4]={1,-1,0,0};
int num[maxn];
int n,m,k,sum,TimeClock;
void dfs(int x,int y){
    Time[x][y]=TimeClock;
    vis[x][y]=1;
    for(int i=0;i<4;i++){
        int xx=x+dx[i];
        int yy=y+dy[i];
        if(xx>=0&&xx<n&&yy>=0&&yy<m&&!vis[xx][yy]){
            if(Map[xx][yy]=='*'){
                sum++;
                Time[xx][yy]=TimeClock;
            }
            if(Map[xx][yy]=='.'){
                dfs(xx,yy);
            }
        }
    }
}
int main()
{
    //freopen("F:\\input.txt","r",stdin);
    while(cin>>n>>m>>k){
        memset(Time,0,sizeof(Time));
        memset(vis,0,sizeof(vis));
        memset(num,0,sizeof(num));
        TimeClock=0;
        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(Map[i][j]=='.'&&!vis[i][j]){
                    sum=0;TimeClock++;
                    dfs(i,j);
                    num[TimeClock]=sum;
                }
            }
        }
        while(k--){
            int wx,wy;
            scanf("%d%d",&wx,&wy);
            printf("%d\n",num[Time[wx-1][wy-1]]);
        }
    }
    return 0;
}