CodeForces 598D Igor In the Museum【dfs】

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 nm 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
8


/*
    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;
}


### Codeforces 1487D Problem Solution The problem described involves determining the maximum amount of a product that can be created from given quantities of ingredients under an idealized production process. For this specific case on Codeforces with problem number 1487D, while direct details about this exact question are not provided here, similar problems often involve resource allocation or limiting reagent type calculations. For instance, when faced with such constraints-based questions where multiple resources contribute to producing one unit of output but at different ratios, finding the bottleneck becomes crucial. In another context related to crafting items using various materials, it was determined that the formula `min(a[0],a[1],a[2]/2,a[3]/7,a[4]/4)` could represent how these limits interact[^1]. However, applying this directly without knowing specifics like what each array element represents in relation to the actual requirements for creating "philosophical stones" as mentioned would require adjustments based upon the precise conditions outlined within 1487D itself. To solve or discuss solutions effectively regarding Codeforces' challenge numbered 1487D: - Carefully read through all aspects presented by the contest organizers. - Identify which ingredient or component acts as the primary constraint towards achieving full capacity utilization. - Implement logic reflecting those relationships accurately; typically involving loops, conditionals, and possibly dynamic programming depending on complexity level required beyond simple minimum value determination across adjusted inputs. ```cpp #include <iostream> #include <vector> using namespace std; int main() { int n; cin >> n; vector<long long> a(n); for(int i=0;i<n;++i){ cin>>a[i]; } // Assuming indices correspond appropriately per problem statement's ratio requirement cout << min({a[0], a[1], a[2]/2LL, a[3]/7LL, a[4]/4LL}) << endl; } ``` --related questions-- 1. How does identifying bottlenecks help optimize algorithms solving constrained optimization problems? 2. What strategies should contestants adopt when translating mathematical formulas into code during competitive coding events? 3. Can you explain why understanding input-output relations is critical before implementing any algorithmic approach? 4. In what ways do prefix-suffix-middle frameworks enhance model training efficiency outside of just tokenization improvements? 5. Why might adjusting sample proportions specifically benefit models designed for tasks requiring both strong linguistic comprehension alongside logical reasoning skills?
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