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.
1 //#pragma comment(linker, "/stack:200000000") 2 //#pragma GCC optimize("Ofast,no-stack-protector") 3 //#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native") 4 //#pragma GCC optimize("unroll-loops") 5 #include<bits/stdc++.h> 6 using namespace std; 7 const int MAX_N = 1e3+9; 8 const int INF = 1e9+7; 9 typedef pair<int,int> P; 10 char vec[MAX_N][MAX_N]; 11 int mp[MAX_N][MAX_N]; 12 int res[MAX_N][MAX_N]; 13 int vis[MAX_N][MAX_N]; 14 int dir[4][2] = {-1,0,1,0,0,1,0,-1}; 15 int ans,flag,N,M,T; 16 queue<P> que; 17 vector<P> st; 18 int dfs(int x,int y,int pos) 19 { 20 if(vis[x][y]) 21 { 22 flag = false; return INF; 23 } 24 if(res[x][y]) return res[x][y]; 25 vis[x][y] = 1; 26 int num =0; 27 for(int i=0;i<4;i++) 28 { 29 int nx = x + dir[i][0]; 30 int ny = y + dir[i][1]; 31 if(nx>=0 && nx<N && ny>=0 && ny<M && mp[nx][ny] == (pos+1)%4 ) 32 { 33 num = max(num,dfs(nx,ny,(pos+1)%4)); 34 } 35 } 36 res[x][y] = num+1; 37 vis[x][y] = 0; 38 return num+1; 39 } 40 int main() 41 { 42 cin>>N>>M; 43 memset(vis,0,sizeof(vis)); 44 memset(res,0,sizeof(res)); 45 st.clear(); 46 for(int i=0; i<N; i++) 47 { 48 scanf("%s",vec[i]); 49 for(int j=0; j<M; j++) 50 { 51 if(vec[i][j] == ‘D‘) mp[i][j] = 0,st.push_back(P(i,j)); 52 else if(vec[i][j] == ‘I‘) mp[i][j] = 1; 53 else if(vec[i][j] == ‘M‘) mp[i][j] = 2; 54 else if(vec[i][j] == ‘A‘) mp[i][j] = 3; 55 } 56 } 57 int out = 0; 58 flag = true; 59 for(int i=0;i<st.size();i++) 60 { 61 if(res[st[i].first][st[i].second] != 0 || vis[st[i].first][st[i].second]) continue; 62 int t = dfs(st[i].first,st[i].second,0); 63 if(!flag) break; 64 out = max(out,t/4); 65 } 66 if(flag) 67 { 68 if(out != 0) cout<<out<<endl; 69 else cout<<"Poor Dima!"<<endl; 70 } 71 else cout<<"Poor Inna!"<<endl; 72 return 0; 73 } 74 /* 75 5 5 76 DIADD 77 DMADD 78 DDDID 79 AMMMD 80 MIDAD 81 82 answer: 3 83 */
