CodeForces - 510B Fox And Two Dots （bfs或dfs）

B. Fox And Two Dots

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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:

1. These k dots are different: if i ≠ j then di is different from dj.
2. k is at least 4.
3. All dots belong to the same color.
4. 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.

Examples

input

Copy

3 4
AAAA
ABCA
AAAA

output

Copy

Yes

input

Copy

3 4
AAAA
ABCA
AADA

output

Copy

No

input

Copy

4 4
YYYR
BYBY
BBBY
BBBY

output

Copy

Yes

input

Copy

7 6
AAAAAB
ABBBAB
ABAAAB
ABABBB
ABAAAB
ABBBAB
AAAAAB

output

Copy

Yes

input

Copy

2 13
ABCDEFGHIJKLM
NOPQRSTUVWXYZ

output

Copy

No

Note

In first sample test all 'A' form a cycle.

In second sample there is no such cycle.

The third sample is displayed on the picture above ('Y' = Yellow, 'B' = Blue, 'R' = Red).

题意：找同一种颜色的环

这题麻烦的地方在于走过的路被标记了，那怎么判断有环呢？

其实记录步数就可以了，bfs和dfs道理是一样的

bfs做法：

#include
using namespace std;
#define ll long long
struct node
{
    int x,y;
};
bool v[55][55];
int book[55][55];
char a[55][55];
int d[4][2]={{-1,0},{1,0},{0,-1},{0,1}};
queueq;
int main()
{
    int n,m;
    cin>>n>>m;
    memset(v,0,sizeof(v));
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=m;j++)
        {
            cin>>a[i][j];
        }
    }
                    bool f=0;
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=m;j++)
        {
            if(!v[i][j])
            {

                memset(book,0,sizeof(book));
                v[i][j]=1;
                while(!q.empty()) q.pop();
                node b;
                b.x=i;
                b.y=j;
                book[i][j]=1;
                q.push(b);
                while(!q.empty())
                {
                    node b=q.front();
                    q.pop();
                    for(int k=0;k<4;k++)
                    {
                        int xx=b.x+d[k][0];
                        int yy=b.y+d[k][1];
                        if(xx<1||yy<1||xx>n||yy>m||a[xx][yy]!=a[i][j]) continue;
                            v[xx][yy]=1;
                            node c;
                            c.x=xx;
                            c.y=yy;
                            if(book[xx][yy]>=book[b.x][b.y])//如果当前走过去格子有步数且的步数比当前这个格子还要大，说明已经走过了，形成了环。
                            {


                                f=1;break;
                            }
                            if(book[xx][yy])continue;
                            q.push(c);
                            book[xx][yy]=book[b.x][b.y]+1;//保存路的步数
                    }
                    if(f) break;
                }
                if(f) break;
            }
            if(f) break;
        }
        if(f) break;
    }
    if(f) cout<<"Yes";
    else cout<<"No";
    return 0;

}

 

dfs也是同样道理：

#include
#include<string>
#include
#include
#include
using namespace std;
char a[55][55];
bool book[55][55];
int v[55][55];
int z[55];
int n,m;
int d[4][2]={{-1,0},{1,0},{0,1},{0,-1}};
bool f=0;
int si,sj;
void dfs(char c,int x,int y)
{
    
    for(int i=0;i<4;i++)
    {
        int xx=x+d[i][0];
        int yy=y+d[i][1];
        if(xx<1||yy<1||xx>n||yy>m) continue;
        if(v[xx][yy]!=0&&v[x][y]-v[xx][yy]>1)//走过去的格子已经有值了且比现在走的格子还大不止1，大1可能是之前走过来的
        {
            f=1;
            return;
        }
        if(a[xx][yy]==c&&v[xx][yy]==0)
        {
            v[xx][yy]=v[x][y]+1;
            book[xx][yy]=1;
            dfs(c,xx,yy);
            if(f) return;
            v[xx][yy]=0;

        }
    }
}

int main()
{
    scanf("%d %d",&n,&m);
    memset(z,0,sizeof(z));
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=m;j++)
        {
            cin>>a[i][j];
        }
    }
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=m;j++)
        {
            if(!book[i][j])
            {
                z[a[i][j]-'A']=1;
                memset(v,0,sizeof(v));
                book[i][j]=1;
                si=i;
                sj=j;
                v[i][j]=1;
                dfs(a[i][j],i,j);
                if(f) break;
            }
        }
        if(f) break;
    }
    if(f) cout<<"Yes";
    else cout<<"No";
    return 0;
}