深度优先搜索

深度优先搜索

今天做出了第一题深度优先搜索题。
至此对广度和深度有了一个基本的了解。
学ACM总算学到了一点非暴力解决问题的方法。
Problem Id:1154  User Id:beyonlin_SCUT
Memory:32K  Time:155MS
Language:C++  Result:Accepted
http://acm.pku.edu.cn/JudgeOnline/problem?id=1154

LETTERS
Time Limit:1000MS  Memory Limit:10000K
Total Submit:694 Accepted:334

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

我的程序：
#include<cstdio>
#include<stack>
using namespace std;
struct node
{
	int row;
	int col;
	int dire;
};
char p[30][30];
char flag[30];
int incr[4][2]={{0,1},{1,0},{0,-1},{-1,0}};
int main()
{
	int i,row,col;
	scanf("%d%d",&row,&col);
	getchar();
	char ch[30];
	for(i=1;i<=row;i++)
	{
		gets(ch);
		int j;
		for(j=1;j<=col;j++)
			p[i][j]=ch[j-1];
	}
	//初始化，外加一层
	for(i=0;i<=col+1;i++)
	{
		p[0][i]='0';
		p[row+1][i]='0';
	}
	for(i=0;i<=row+1;i++)
	{
		p[i][0]='0';
		p[i][col+1]='0';
	}
	int Maxmove=0;//最大步数
	stack<node>path;
        //栈初始化
	int r=1,c=1,dire=0,f=0,move=1;
	node in;
	in.row=r;
	in.col=c;
	in.dire=dire;
	path.push(in);
	flag[f++]=p[r][c];
	while(!path.empty())
	{
		if(dire<4)
		{
			int r2=r+incr[dire][0];
			int c2=c+incr[dire][1];
			bool b=true;
			for(int k=0;k<f;k++)//搜索是否已访问或路不通
			{
				if(flag[k]==p[r2][c2] || p[r2][c2]=='0')
				{
					dire++;
					b=false;
					break;
				}
			}
			if(b)//路通
			{
				node in;
				in.row=r2;
				in.col=c2;
				in.dire=dire;
				path.push(in);//进栈
				move++;
				flag[f++]=p[r2][c2];//标志已访问
				r=r2;
				c=c2;
				dire=0;
			}
		}
		else//找到一个解
		{
			if(move>Maxmove)
				Maxmove=move;
			move--;
			dire=path.top().dire+1;	//回溯，去除访问标志
			path.pop();
			flag[--f]='\0';
			if(!path.empty())
			{
				r=path.top().row;
				c=path.top().col;
			}
		}
	}
	printf("%d\n",Maxmove);
	return 0;
}