1 LETTERS
2 Time Limit: 1000MS Memory Limit: 10000K
3 Total Submissions: 9298 Accepted: 4174
4 Description
5
6 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.
7 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.
8 The goal of the game is to play as many moves as possible.
9 Write a program that will calculate the maximal number of positions in the board the figure can visit in a single game.
10 Input
11
12 The first line of the input contains two integers R and C, separated by a single blank character, 1 <= R, S <= 20.
13 The following R lines contain S characters each. Each line represents one row in the board.
14 Output
15
16 The first and only line of the output should contain the maximal number of position in the board the figure can visit.
17 Sample Input
18
19 3 6
20 HFDFFB
21 AJHGDH
22 DGAGEH
23 Sample Output
24
25 6
1 #include <iostream>
2 #include <algorithm>
3 #include <cstdio>
4 #include <string>
5 #include <string>
6 #include <cstring>
7 #include <map>
8 #include <utility>
9 using namespace std;
10 const int N = 1e2+20 ;
11 int n,m;
12 int maxx,cnt;
13 bool vis[N];
14 char s[N][N];
15 int dir[4][2]={
16 {0,1},{0,-1},{-1,0},{1,0}
17 };
18 bool check(int r,int l){
19 if(r>=0&&r<n&&l>=0&&l<m&&vis[s[r][l]-'A']==0){
20 return true;
21 }
22 return false;
23 }
24 void dfs(int r,int l){
25
26
27 maxx=max(maxx,cnt);
28 vis[s[r][l]-'A']=1;
29 for(int i=0;i<4;i++){
30 int rr=r+dir[i][0];
31 int ll=l+dir[i][1];
32 if(check(rr,ll)){
33 cnt++;
34 dfs(rr,ll);
35 vis[s[rr][ll]-'A']=0;
36 cnt--;
37 }
38 }
39 }
40 int main()
41 {
42 while (~scanf("%d%d",&n,&m)){
43 memset(vis,0,sizeof(vis));
44 for(int i=0;i<n;i++) scanf("%s",s[i]);
45 maxx=1;cnt=1;
46 dfs(0,0);
47 printf("%d\n",maxx);
48 }
49 return 0;
50 }
