dfs(最长路径)

http://poj.org/problem?id=1154

LETTERS

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 9519		Accepted: 4252

Description

A single-player game is played on a rectangular board divided in R rows and C columns. There is a single uppercase letter (A-Z) written in every position in the board.
Before the begging of the game there is a figure in the upper-left
corner of the board (first row, first column). In every move, a player
can move the figure to the one of the adjacent positions (up, down,left
or right). Only constraint is that a figure cannot visit a position
marked with the same letter twice.

The goal of the game is to play as many moves as possible.

Write a program that will calculate the maximal number of positions in the board the figure can visit in a single game.

Input

The first line of the input contains two integers R and C, separated by a single blank character, 1 <= R, S <= 20.

The following R lines contain S characters each. Each line represents one row in the board.

Output

The first and only line of the output should contain the maximal number of position in the board the figure can visit.

Sample Input

3 6
HFDFFB
AJHGDH
DGAGEH

Sample Output

6

Source

Croatia OI 2002 Regional Competition - Juniors

题意：从左上角出发，每个字母不走两次的条件下，求走最多次数。

//#include <bits/stdc++.h>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <algorithm>
#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <stdio.h>
#include <queue>
#include <stack>;
#include <map>
#include <set>
#include <string.h>
#include <vector>
#define ME(x , y) memset(x , y , sizeof(x))
#define SF(n) scanf("%d" , &n)
#define rep(i , n) for(int i = 0 ; i < n ; i ++)
#define INF  0x3f3f3f3f
#define mod 1000000007
using namespace std;
typedef long long ll ;
char a[][];
int n , m ;
int bx , by ;
int ans ;
int vis[];
int dir[][] = {{ , } , {- , } , { , } , { , -}};

void dfs(int x , int y , int cnt)
{
    for(int i =  ; i <  ; i++)
    {

        int xx = x + dir[i][];
        int yy = y + dir[i][];
        if(vis[a[xx][yy] - 'A'] || xx <=  || xx > n || yy <=  || yy > m)
        {
            continue;
        }
        vis[a[xx][yy] - 'A'] =  ;
        dfs(xx , yy , cnt+);

    }
    ans = max(ans , cnt);//for循环结束代表一条路径的结束。
    vis[a[x][y] - 'A'] =  ;//不同路径互不干扰
}

int main()
{
    while(~scanf("%d%d" , &n , &m))
    {
        for(int i =  ; i <= n ; i++)
        {
            for(int j =  ; j <= m ; j++)
            {
                cin >> a[i][j];
            }
        }
        memset(vis ,  , sizeof(vis));
        vis[a[][] - 'A'] =  ;//将原地标记并加一
        ans =  ;
        dfs( ,  , );
        cout << ans << endl ;
    }

    return ;
}

下面这个是参考学长的码。

//#include <bits/stdc++.h>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <algorithm>
#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <stdio.h>
#include <queue>
#include <stack>;
#include <map>
#include <set>
#include <string.h>
#include <vector>
#define ME(x , y) memset(x , y , sizeof(x))
#define SF(n) scanf("%d" , &n)
#define rep(i , n) for(int i = 0 ; i < n ; i ++)
#define INF  0x3f3f3f3f
#define mod 1000000007
using namespace std;
typedef long long ll ;
char a[][];
int n , m ;
int ans ;
int vis[];
int dir[][] = {{ , } , {- , } , { , } , { , -}};

void dfs(int x , int y , int cnt)
{
    if(vis[a[x][y] - 'A'] || x <=  || y <=  || x > n || y > m)
    {
        cnt--;
        return ;
    }
    else vis[a[x][y] - 'A'] =  ;
    dfs(x+ , y , cnt+);
    dfs(x- , y , cnt+);
    dfs(x , y+ , cnt+);
    dfs(x , y- , cnt+);
    ans = max(cnt , ans);
    vis[a[x][y] - 'A'] =  ;//每一条走到最后的路径都互不影响，所以要取消标记。
}

int main()
{
    while(~scanf("%d%d" , &n , &m))
    {
        for(int i =  ; i <= n ; i++)
        {
            for(int j =  ; j <= m ; j++)
            {
                cin >> a[i][j];
            }
        }
        ans = -INF ;
        memset(vis ,  , sizeof(vis));
        dfs( ,  , );
        cout << ans << endl ;
    }

    return ;
}