LETTERS

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 8976		Accepted: 4017

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.

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.

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

3 6

HFDFFB

AJHGDH

DGAGEH

6
思路： 
基础dfs，
实现代码：

//#include<bits/stdc++.h>

#include<iostream>

#include<cstdio>

#include<string>

#include<cstring>

#include<cmath>

#include<algorithm>

#include<map>

#include<queue>

#include<stack>

#include<set>

#include<list>

using namespace std;

#define ll long long

const int Mod = 1e9+;

const int inf = 1e9;

const int Max = 1e5+;

vector<int>vt[Max];

int dx[] = {-, ,  , };

int dy[] = { , , -, };

//void exgcd(ll a,ll b,ll& d,ll& x,ll& y){if(!b){d=a;x=1;y=0;}else{exgcd(b,a%b,d,y,x);y-=x*(a/b);}}

//ll inv(ll a,ll n){ll d, x, y;exgcd(a,n,d,x,y);return (x+n)%n;}  ��Ԫ

//int gcd(int a,int b)  {  return (b>0)?gcd(b,a%b):a;  }  ��С��Լ

//int lcm(int a, int b)  {  return a*b/gcd(a, b);   }    ������

int ans = ,n,m,vis[];

char mp[][];

void dfs(int x,int y,int cnt){

    for(int i=;i<;i++){

        int nx = dx[i] + x;

        int ny = dy[i] + y;

        if(nx>=&&nx<n&&ny>=&&ny<m&&vis[mp[nx][ny]-'A']==){

            vis[mp[nx][ny]-'A'] = ;

            ans = max(ans,cnt+);

            dfs(nx,ny,cnt+);

            vis[mp[nx][ny]-'A'] = ;

        }

    }

}

int main()

{

    cin>>n>>m;

    memset(vis,,sizeof(vis));

    for(int i=;i<n;i++){

        for(int j=;j<m;j++){

            cin>>mp[i][j];

        }

    }

    vis[mp[][]-'A'] = ;

    dfs(,,);

    cout<<ans<<endl;

}