Description
You have a maze with obstacles and non-zero digits in it:
You can start from any square, walk in the maze, and finally stop at some square. Each step, you may only walk into one of the four neighbouring squares (up, down, left, right) and you cannot walk into obstacles or walk into a square more than once. When you finish, you can get a number by writing down the digits you encounter in the same order as you meet them. For example, you can get numbers 9784, 4832145 etc. The biggest number you can get is 791452384, shown in the picture above.
Your task is to find the biggest number you can get.
Input
There will be at most 25 test cases. Each test begins with two integers R and C (2≤R,C≤15, R∗C≤30), the number of rows and columns of the maze. The next R rows represent the maze. Each line contains exactly C characters (without leading or trailing spaces), each of them will be either ‘#’ or one of the nine non-zero digits. There will be at least one non-obstacle squares (i.e. squares with a non-zero digit in it) in the maze. The input is terminated by a test case with R=C=0, you should not process it.
Output
For each test case, print the biggest number you can find, on a single line.
Samples
Input
3 7 ##9784# ##123## ##45### 0 0
Output
791452384
在训练的时候直接硬搜，TLE
TLE_code:
int n,m; vector<string>vet[33]; char s[33][33]; bool vis[30][30]; bool edge(int x,int y) { if(x < 1 || x > n || y < 1 || y > m) return false; return true; } int dx[5]= {0,0,0,1,-1}; int dy[5]= {0,1,-1,0,0}; void dfs(int x,int y,int len,string ans) { if(!edge(x,y)) return; int flag = 0; for(int i=1; i<=4; i++) { int tx = x + dx[i]; int ty = y + dy[i]; if(edge(tx,ty) && s[tx][ty] !='#' && !vis[tx][ty]) { vis[x][y] = 1; flag = 1; dfs(tx,ty,len+1,ans+s[x][y]); vis[x][y] = 0; } } if(!flag) vet[len+1].push_back(ans+s[x][y]); } int main() { while(cin >> n >> m && n && m) { for(int i=1; i<=n; i++) cin >> (s[i] + 1); for(int i=0; i<33; i++) vet[i].clear(); for(int i=1; i<=n; i++) { for(int j=1; j<=m; j++) { if(isdigit(s[i][j])) { ///memset(vis,0,sizeof vis); vis[i][j] = 1; dfs(i,j,0,""); vis[i][j] = 0; } } } for(itn i=30; i>=1; i--) { if(vet[i].size()) { sort(vet[i].begin(),vet[i].end()); int siz = vet[i].size(); printf("%s\n",vet[i][siz-1].c_str()); /// cout << vet[i][siz-1] <<endl; break; } } } return 0; } /** 3 7 ##9784# ##123## ##45### **/
以上代码是将所有的结果都进行保存放到vector里，然后按照长度从大到小进行查找符合而条件的最大的一个字符串
/*** keep hungry and calm PushyTao!***/ typedef int itn; int n,m,flag,mx; char s[33][33]; bool vis[30][30]; char a[33][33]; char ans[40],cur[40]; bool edge(int x,int y) { if(x < 1 || x > n || y < 1 || y > m) return false; return true; } int dx[5]= {0,0,0,1,-1}; int dy[5]= {0,1,-1,0,0}; bool cmp(int len){ if(!strlen(ans)) return true; for(int i=0;i<len;i++){ if(ans[i] != cur[i]) return ans[i] < cur[i]; } return true; } void dfs(int x,int y,int len){ if(len == mx){ flag = 1; if(cmp(len)) strcpy(ans,cur),ans[len]='\0'; return; } for(itn i=1;i<=4;i++){ int tx = x + dx[i]; int ty = y + dy[i]; if(edge(tx,ty) && a[tx][ty] && !vis[tx][ty]){ cur[len] = a[tx][ty]; if(cmp(len)){ ///cur[len] = a[tx][ty]; vis[tx][ty] = 1; dfs(tx,ty,len+1); vis[tx][ty] = 0; } } } } int main() { while(cin >> n >> m && n && m) { memset(a,0,sizeof a); memset(vis,0,sizeof vis); memset(ans,0,sizeof ans); flag = 0; int cnt = 0; for(int i=1;i<=n;i++){ cin >> (s[i] + 1); for(int j=1;j<=m;j++){ if(s[i][j] != '#') a[i][j] = s[i][j],cnt++; } } for(mx = cnt;!flag;mx--){ for(itn i=1;i<=n;i++){ for(int j=1;j<=m;j++){ if(a[i][j]){ cur[0] = a[i][j]; vis[i][j] = 1; dfs(i,j,1); vis[i][j] = 0; } } } } cout << ans <<endl; } return 0; }
