CF Fox And Two Dots (DFS)

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.

Sample test(s)

input

3 4
AAAA
ABCA
AAAA

output

Yes

input

3 4
AAAA
ABCA
AADA

output

No

input

4 4
YYYR
BYBY
BBBY
BBBY

output

Yes

input

7 6
AAAAAB
ABBBAB
ABAAAB
ABABBB
ABAAAB
ABBBAB
AAAAAB

output

Yes

input

2 13
ABCDEFGHIJKLM
NOPQRSTUVWXYZ

output

No

读错题了，一直以为是要找矩形，比赛快结束才被队友提醒。
DFS一下就可以，每个点记录一下它是从起点出发的第几个点，如果搜索的过程中遇到了走过的点，那么判断一下这两个点的差是否大于等于3，如果满足就说明形成了一个环。本来想的是，如果从某个点出发没找到，那么就说明和这个点连通的所有点都不可行，于是加入了一个剪枝，将这一连通分量减掉，但是似乎没什么作用，加和不加都是15ms，网上还有一种更炫的写法，就是把起点放在当前点的屁股后面，如果遇到了走过的点就找到，我有空试试，写出来的话就更新上来。

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深夜补发，刚才说的那个算法写了下，实际效果没想象的那么好，还没有我之前写的那个快，不过仔细想想，复杂度貌似是一样的。

不加剪枝：

#include <bits/stdc++.h>
using    namespace    std;

const    int    UPDATE[][] = {{,},{,-},{,},{-,}};
char    MAP[][];
int    N,M;
struct    Node
{
    bool    vis = ;
    int    n = ;
}VIS[][];

bool    dfs(int,int,int,char);
int    main(void)
{
    cin >> N >> M;

    for(int i = ;i <= N;i ++)
        cin >> MAP[i] + ;
    for(int i = ;i <= N;i ++)
        for(int j = ;j <= M;j ++)
        {
            VIS[i][j].vis = VIS[i][j].n = ;
            if(dfs(i,j,,MAP[i][j]))
            {
                puts("Yes");
                return    ;
            }
            VIS[i][j].vis = VIS[i][j].n = ;
        }
    puts("No");

    return    ;
}

bool    dfs(int x,int y,int sum,char color)
{
    for(int i = ;i < ;i ++)
    {
        int    new_x = x + UPDATE[i][];
        int    new_y = y + UPDATE[i][];
        if(new_x > N || new_x <  || new_y > M || new_y <  ||
           MAP[new_x][new_y] != color)
            continue;

        if(VIS[new_x][new_y].vis && sum - VIS[new_x][new_y].n +  >= )
            return    true;
        if(VIS[new_x][new_y].vis)
            continue;

        VIS[new_x][new_y].vis = true;
        VIS[new_x][new_y].n = sum + ;
        if(dfs(new_x,new_y,sum + ,color))
            return    true;
        VIS[new_x][new_y].vis = false;
        VIS[new_x][new_y].n = sum;
    }

    return    false;
}

加了剪枝:

#include <bits/stdc++.h>
using    namespace    std;

const    int    UPDATE[][] = {{,},{,-},{,},{-,}};
char    MAP[][];
int    N,M;
struct    Node
{
    bool    vis = ;
    int    n = ;
}VIS[][];

bool    dfs(int,int,int,char);
void    cut(int,int,char);
int    main(void)
{
    cin >> N >> M;

    for(int i = ;i <= N;i ++)
        cin >> MAP[i] + ;
    for(int i = ;i <= N;i ++)
        for(int j = ;j <= M;j ++)
        {
            if(MAP[i][j] == '.')
                continue;
            VIS[i][j].vis = VIS[i][j].n = ;
            if(dfs(i,j,,MAP[i][j]))
            {
                puts("Yes");
                return    ;
            }
            VIS[i][j].vis = VIS[i][j].n = ;
            cut(i,j,MAP[i][j]);
        }
    puts("No");

    return    ;
}

bool    dfs(int x,int y,int sum,char color)
{
    for(int i = ;i < ;i ++)
    {
        int    new_x = x + UPDATE[i][];
        int    new_y = y + UPDATE[i][];
        if(new_x > N || new_x <  || new_y > M || new_y <  ||
           MAP[new_x][new_y] != color)
            continue;

        if(VIS[new_x][new_y].vis && sum - VIS[new_x][new_y].n +  >= )
            return    true;
        if(VIS[new_x][new_y].vis)
            continue;

        VIS[new_x][new_y].vis = true;
        VIS[new_x][new_y].n = sum + ;
        if(dfs(new_x,new_y,sum + ,color))
            return    true;
        VIS[new_x][new_y].vis = false;
        VIS[new_x][new_y].n = sum;
    }

    return    false;
}

void    cut(int x,int y,char cor)
{
    for(int i = ;i < ;i ++)
    {
        int    new_x = x + UPDATE[i][];
        int    new_y = y + UPDATE[i][];
        if(new_x > N || new_x <  || new_y > M || new_y <  ||
           MAP[new_x][new_y] != cor || VIS[new_x][new_y].vis)
            continue;
        MAP[new_x][new_y] = '.';
        cut(new_x,new_y,cor);
    }
}

网上的算法:

#include <bits/stdc++.h>
using    namespace    std;

const    int    UPDATE[][] = {{,},{,-},{,},{-,}};
int    N,M;
char    MAP[][];
bool    VIS[][];
int    BACK_X,BACK_Y;

bool    dfs(int i,int j,char color);
int    main(void)
{
    cin >> N >> M;
    for(int i = ;i <= N;i ++)
        cin >> MAP[i] + ;
    for(int i = ;i <= N;i ++)
        for(int j = ;j <= M;j ++)
        {
            VIS[i][j] = ;
            if(dfs(i,j,MAP[i][j]))
            {
                puts("Yes");
                return    ;
            }
            VIS[i][j] = ;
        }
    puts("No");

    return    ;
}

bool    dfs(int i,int j,char color)
{
    int    back_x = BACK_X;
    int    back_y = BACK_Y;
    for(int k = ;k < ;k ++)
    {
        int    next_x = i + UPDATE[k][];
        int    next_y = j + UPDATE[k][];

        if(next_x > N || next_x <  || next_y > M || next_y <  ||
           MAP[next_x][next_y] != color)
            continue;
        if(VIS[next_x][next_y] && next_x != BACK_X && next_y != BACK_Y)
        {
            return    true;
        }
        if(VIS[next_x][next_y])
            continue;

        BACK_X = i;
        BACK_Y = j;
        VIS[next_x][next_y] = ;
        if(dfs(next_x,next_y,color))
            return    true;
        VIS[next_x][next_y] = ;
        BACK_X = back_x;
        BACK_Y = back_y;
    }

    return    false;
}