POJ1154


LETTERS

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 8119		Accepted: 3661

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
题意：
给出一个大写字母矩阵，一开始位于左上角，可以上下左右移动但不能移动到曾经经过的字母，问最多可以经过几个字母。
思路：

基础DFS，以前做这道题时，总是不理解怎么统计经过字母的个数和怎么让vis返回，现在重新做终于可以自己理解并解决了。
代码：
#include<iostream>

#include<string>

#include<cstdio>

#include<cmath>

#include<cstring>

using namespace std;

int r,s,sum,cnt,dir[4][2]={{1,0},{0,1},{-1,0},{0,-1}};

bool vis[30];

char map[30][30];

void dfs(int x,int y)

{

    if(cnt>sum) sum=cnt;

    vis[map[x][y]-'A']=1;

    for(int i=0;i<4;i++)

    {

        int x1=x+dir[i][0];

        int y1=y+dir[i][1];

        if(x1>=1&&x1<=r&&y1>=1&&y1<=s&&!vis[map[x1][y1]-'A'])

        {

            cnt++;

            dfs(x1,y1);

            vis[map[x1][y1]-'A']=0;

            cnt--;

        }

    }

}

int main()

{

    while(cin>>r>>s)

    {

        for(int i=1;i<=r;i++)

        {

            for(int j=1;j<=s;j++)

            cin>>map[i][j];

        }

        memset(vis,0,sizeof(vis));

        sum=1;cnt=1;

        dfs(1,1);

        cout<<sum<<endl;

    }

    return 0;

}