CodeForces-510B Fox And Two Dots

传送门：CodeForces - 510B

题目描述

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 $d_1$, $d_2$, …, $d_k$ a cycle if and only if it meets the following condition:

1. These k dots are different: if i ≠ j then $d_i$ is different from $d_j$.
2. k is at least 4.
3. All dots belong to the same color.
4. For all $1 ≤ i ≤ k - 1$: $d_i$ and $d_i$ + 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.

输入描述

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.

输出描述

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

题意分析

在一个n*m的矩阵中，能否用同一种字母组成一个环，从起点点开始上下左右，并且将途经所有点标记，如果最后能回到起点，说明能组成一个环，要注意一点的就是，搜索过程中下一个点不能与上一个点相同

样例输入

3 4
AAAA
ABCA
AAAA

样例输出

Yes

AC代码

#include<bits/stdc++.h>
#pragma GCC optimize(2)

using namespace std;

typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef map<int, int> mii;
typedef map<ll, ll> mll;

#define ull unsigned long long
#define IOS ios_base::sync_with_stdio(0); cin.tie(0);cout.tie(0);

const int maxn=1e6+10;
const ll mod=1e9+7;
const int INF = 0x3f3f3f3f;

int n,m;
char a[55][55];
int vis[55][55];
int d1[4]= {1,-1,0,0},d2[4]= {0,0,1,-1};
int f=0;

bool judge(int x,int y) {
    if(x<0||y<0||x>=n||y>=m)return false;
    else return true;
}

void dfs(int x,int y,int xx,int yy) {
    if(vis[x][y]) {
        f=1;
        return;
    }
    vis[x][y]=1;
    for(int i=0; i<4; i++) {
        int nx,ny;
        nx=x+d1[i];
        ny=y+d2[i];
        if(judge(nx,ny) && a[nx][ny]==a[x][y] && (nx!=xx||ny!=yy) ) {
            dfs(nx,ny,x,y);
        }
    }

}

int main() {
    IOS;
    int i,j;
    cin>>n>>m;
    for(i=0; i<n; i++) {
        for(j=0; j<m; j++) {
            cin>>a[i][j];
        }
    }
    for(i=0; i<n; i++) {
        for(j=0; j<m; j++) {
            if(!vis[i][j]) {
                dfs(i,j,i,j);
                if(f)break;
            }
        }
    }
    if(f)cout<<"Yes"<<endl;
    else cout<<"No"<<endl;

}