Codeforces 374 C. Travelling Salesman and Special Numbers (dfs、记忆化搜索)

题目链接：Travelling Salesman and Special Numbers

题意：

　　给了一个n×m的图，图里面有'N','I','M','A'四种字符。问图中能构成NIMA这种序列最大个数(连续的，比如说NIMANIMA = 2)为多少，如果有环的话那么最大长度就是无穷。

题解：

　　哇，做了这题深深得感觉自己的dfs真的好弱。这题就是从N开始深搜，在深搜的过程中记录值。返回这个值加1.

 //#pragma comment(linker, "/stack:200000000")
 //#pragma GCC optimize("Ofast,no-stack-protector")
 //#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
 //#pragma GCC optimize("unroll-loops")
 #include<bits/stdc++.h>
 using namespace std;
 const int MAX_N = 1e3+;
 const int INF = 1e9+;
 typedef pair<int,int> P;
 char vec[MAX_N][MAX_N];
 int mp[MAX_N][MAX_N];
 int res[MAX_N][MAX_N];
 int vis[MAX_N][MAX_N];
 int dir[][] = {-,,,,,,,-};
 int ans,flag,N,M,T;
 queue<P> que;
 vector<P> st;
 int dfs(int x,int y,int pos)
 {
     if(vis[x][y])
     {
         flag = false; return INF;
     }
     if(res[x][y]) return res[x][y];
     vis[x][y] = ;
     int num =;
     for(int i=;i<;i++)
     {
         int nx = x + dir[i][];
         int ny = y + dir[i][];
         if(nx>= && nx<N && ny>= && ny<M && mp[nx][ny] == (pos+)% )
         {
             num = max(num,dfs(nx,ny,(pos+)%));
         }
     }
     res[x][y] = num+;
     vis[x][y] = ;
     return num+;
 }
 int main()
 {
     cin>>N>>M;
     memset(vis,,sizeof(vis));
     memset(res,,sizeof(res));
     st.clear();
     for(int i=; i<N; i++)
     {
         scanf("%s",vec[i]);
         for(int j=; j<M; j++)
         {
             if(vec[i][j] == 'D') mp[i][j] = ,st.push_back(P(i,j));
             else if(vec[i][j] == 'I') mp[i][j] = ;
             else if(vec[i][j] == 'M') mp[i][j] = ;
             else if(vec[i][j] == 'A') mp[i][j] = ;
         }
     }
     int out = ;
     flag = true;
     for(int i=;i<st.size();i++)
     {
         if(res[st[i].first][st[i].second] !=  || vis[st[i].first][st[i].second]) continue;
         int t = dfs(st[i].first,st[i].second,);
         if(!flag) break;
         out = max(out,t/);
     }
     if(flag)
     {
         if(out != ) cout<<out<<endl;
         else cout<<"Poor Dima!"<<endl;
     }
     else cout<<"Poor Inna!"<<endl;
     return ;
 }
 /*
 5 5
 DIADD
 DMADD
 DDDID
 AMMMD
 MIDAD

 answer: 3
 */