1154:LETTERS

题目链接http://bailian.openjudge.cn/practice/1154/

总时间限制: 1000ms 内存限制: 65536kB

描述

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.

输入

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.

输出

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

样例输入

3 6
HFDFFB
AJHGDH
DGAGEH

样例输出

6

来源

Croatia OI 2002 Regional Competition - Juniors

算法：深搜

代码一：

#include<iostream>
using namespace std;
int bb[26]={0},s,r,sum=1,s1=1;
char aa[25][25];
int dir[4][2]={-1,0,1,0,0,-1,0,1};
void dfs(int a,int b)
{
    int a1,b1;
    if(s1>sum) sum=s1;           //更新最大数值
    for(int i=0;i<4;i++)
    {
        a1=a+dir[i][0];             //用bb数组记录访问过的字母
        b1=b+dir[i][1];
        if(a1>=0&&a1<s&&b1>=0&&b1<r&&!bb[aa[a1][b1]-'A'])
        {
            s1++;
            bb[aa[a1][b1]-'A']=1;   //如果在这条单线上没有记录改字母被访问过，则总数++；
            dfs(a1,b1);             //第一个字母总要被访问，所以不用回溯；
            bb[aa[a1][b1]-'A']=0;   //回溯反标记
            s1--;                   //临时记录恢复
        }
    }
}
int main()
{
    cin>>s>>r;
    for(int i=0;i<s;i++)
        for(int j=0;j<r;j++)
        cin>>aa[i][j];
    bb[aa[0][0]-'A']=1;
    dfs(0,0);
    cout<<sum<<endl;
    return 0;
}

代码二：

#include <stdio.h>
#include<iostream>
using namespace std;
int qq[25][25];
int fx[4]={1,0,-1,0},fy[4]={0,-1,0,1},pd[30],sum,ans;//右下左上 
void fun(int x,int y)
{    
    if(ans<sum)ans=sum;
    if(qq[x][y]==0) return;
    for(int i=0;i<4;i++)
    {
        if(qq[x+fx[i]][y+fy[i]]!=0&&pd[qq[x+fx[i]][y+fy[i]]]==0)
        {
            sum++;
            pd[qq[x+fx[i]][y+fy[i]]]=1;
            fun(x+fx[i],y+fy[i]);
            pd[qq[x+fx[i]][y+fy[i]]]=0;
            sum--;
        }
    }
}
int main(int argc, char *argv[])
{
    int r,s;
    scanf("%d%d",&r,&s);
    for(int i=1;i<=r;i++)
        for(int j=1;j<=s;j++)
           {
            char t;
            cin>>t; 
            qq[i][j]=t-'A'+1;
        }
    pd[qq[1][1]]=1;
    sum=ans=1;
    fun(1,1);
    printf("%d",ans);
    return 0;
}