• Codeforces Round #290 (Div. 2) B. Fox And Two Dots dfs


    B. Fox And Two Dots

    题目连接:

    http://codeforces.com/contest/510/problem/B

    Description

    Fox Ciel is playing a mobile puzzle game called "Two Dots". The basic levels are played on a board of size n × m cells, like this:

    Each cell contains a dot that has some color. We will use different uppercase Latin characters to express different colors.

    The key of this game is to find a cycle that contain dots of same color. Consider 4 blue dots on the picture forming a circle as an example. Formally, we call a sequence of dots d1, d2, ..., dk a cycle if and only if it meets the following condition:

    These k dots are different: if i ≠ j then di is different from dj.

    k is at least 4.

    All dots belong to the same color.

    For all 1 ≤ i ≤ k - 1: di and di + 1 are adjacent. Also, dk and d1 should also be adjacent. Cells x and y are called adjacent if they share an edge.

    Determine if there exists a cycle on the field.

    Input

    The first line contains two integers n and m (2 ≤ n, m ≤ 50): the number of rows and columns of the board.

    Then n lines follow, each line contains a string consisting of m characters, expressing colors of dots in each line. Each character is an uppercase Latin letter.

    Output

    Output "Yes" if there exists a cycle, and "No" otherwise.

    Sample Input

    3 4
    AAAA
    ABCA
    AAAA

    Sample Output

    Yes

    Hint

    题意

    给你一个n*m的网格

    然后问你是否有只含有一种元素的环

    题解:

    dfs就好了

    dfs的时候,记录一下fa,然后一直跑下去,跑到曾经vis过的地方,就说明遇到了环

    代码

    #include<bits/stdc++.h>
    using namespace std;
    
    char mp[55][55];
    int vis[55][55];
    int flag = 0;
    int dx[4]={1,-1,0,0};
    int dy[4]={0,0,1,-1};
    int n,m;
    void dfs(int x,int y,char c,int fax,int fay)
    {
        vis[x][y]=1;
        if(flag)return;
        for(int i=0;i<4;i++)
        {
            int xx = x+dx[i];
            int yy = y+dy[i];
            if(xx==fax&&yy==fay)continue;
            if(xx<0||xx>=n)continue;
            if(yy<0||yy>=m)continue;
            if(mp[xx][yy]!=c)continue;
            if(vis[xx][yy]){
                flag=1;
                return;
            }
            dfs(xx,yy,c,x,y);
        }
    }
    int main()
    {
        scanf("%d%d",&n,&m);
        for(int i=0;i<n;i++)
            scanf("%s",&mp[i]);
        for(int i=0;i<n;i++)
            for(int j=0;j<m;j++)
            {
                if(!vis[i][j])
                    dfs(i,j,mp[i][j],i,j);
            }
        if(flag)printf("Yes");
        else printf("No
    ");
    }
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  • 原文地址:https://www.cnblogs.com/qscqesze/p/5077090.html
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