Roughly meaning of the title is A fancy B, A to C by a friend of the same sex, if C know D, and D is the same sex friend B, then B of A can pursue.
Method One: DFS violence enumeration method, cool ~ ~.
1 #include<iostream> 2 #include<vector> 3 #include<unordered_map> 4 #include<algorithm> 5 using namespace std; 6 7 struct node { 8 int a,b; 9 }; 10 const int maxn = 310; 11 int n,m,k; 12 13 vector<int> G[maxn],path; 14 bool visited[maxn] = {false}; 15 vector<node> ans; 16 bool cmp(node& a,node& b) { 17 if(a.a != b.a) return a.a < b.a; 18 else return a.b < b.b; 19 } 20 unordered_map<string,int> IDtoInt; 21 unordered_map<int,string> IntToID; 22 int num = 1; 23 int getID(string id) { 24 if(IDtoInt.count(id) == 0) { 25 IDtoInt[id] = num; 26 IntToID[num++] = id; 27 } 28 return IDtoInt[id]; 29 } 30 31 void DFS(int now,int end,int len) { 32 if(len == 4) { 33 if(now == end) { 34 path.push_back(end); 35 if(IntToID[path[0]].size() == IntToID[path[1]].size() &&IntToID[path[2]].size() == IntToID[path[3]].size() ) 36 ans.push_back({abs(stoi(IntToID[path[1]])),abs(stoi(IntToID[path[2]]))}); 37 path.pop_back(); 38 } 39 return ; 40 } 41 if(now == end && len < 4) return; 42 visited[now] = true; 43 path.push_back(now); 44 for(int i = 0; i < G[now].size(); ++i) { 45 if(visited[G[now][i]] == false) 46 DFS(G[now][i],end,len+1); 47 } 48 visited[now] = false; 49 path.pop_back(); 50 } 51 int main() { 52 scanf("%d%d",&n,&m); 53 string a,b; //必须考虑 -0000,0000的情况，所以用string 54 for(int i = 0; i < m; ++i) { 55 cin>>a>>b; 56 int u = getID(a),v = getID(b); 57 G[u].push_back(v); 58 G[v].push_back(u); 59 } 60 scanf("%d",&k); 61 while(k--) { 62 cin>>a>>b; 63 ans.clear(); 64 DFS(IDtoInt[a],IDtoInt[b],1); 65 sort(ans.begin(),ans.end(),cmp); 66 printf("%d\n",ans.size()); 67 for(int i = 0; i < ans.size(); ++i) printf("%04d %04d\n",ans[i].a,ans[i].b); 68 } 69 return 0; 70 }
Method Two:
Establishing a corresponding relationship with the ID number with the map.
Adjacency table holds a pair of vertices of friends.
Adjacency matrix, all save a person's same-sex friends.
If same-sex friends A is C, B is the same-sex friend D, if C and D is a friend, then A can pursue B.
important point:
1, the number must be from 1 to start, because the ID given possible illegal, the corresponding number of 0 indicates not find this person, or test point 5 can not pass.
2, pay attention to the input as a -0000, -0000 must be accepted with string
1 #include<iostream> 2 #include<vector> 3 #include<algorithm> 4 #include<unordered_map> 5 using namespace std; 6 const int maxn = 320; 7 struct node { 8 int a,b; 9 }; 10 int n,m,k,G[maxn][maxn]; 11 vector<int> adj[maxn]; 12 bool cmp(node a,node b) { 13 if(a.a != b.a) return a.a < b.a; 14 the else return ab & < BB; 15 } 16 unordered_map < String , int > IDtoInt; . 17 unordered_map < int , String > IntToID; 18 is int NUM = . 1 ; // write num = 0, the lead test points 5 can not pass, because the input illicit IDtoInt [ID] = 0, so the number num from a legitimate start of ID . 19 int the getID ( String ID) { 20 is IF (! IDtoInt.count (ID)) { 21 is IDtoInt [ID] = num; 22 is IntToID [num ++] = the above mentioned id; 23 } 24 return IDtoInt [ID]; 25 } 26 is int main () { 27 Fill (G [ 0 ], G [ 0 ] + MAXN MAXN *, 0 ); 28 CIN >> >> n- m; 29 String A, B ; 30 for ( int I = 0 ; I <m; ++ I) { 31 is CIN >> a >> B; // the reason for this is that there are error-prone reading point: -0000 32 int U = the getID (a) , V = the getID (B); 33 is G [U] [V] = G [V] [U] = . 1 ; // friendship 34 IF (a.size () == b.size ()) { // sex friend relationship 35 ADJ [U] .push_back (V); 36 ADJ [V] .push_back (U); 37 [ } 38 is } 39 CIN> > K; 40 the while (K-- ) { 41 is CIN >> A >> B; 42 is Vector <Node> ANS; 43 is int U = IDtoInt [A], V = IDtoInt [B]; 44 is for ( int I = 0 ; I <ADJ [u] .size (); I ++) { // find u pals C 45 int C = adj[u][i]; 46 for(int j = 0; j < adj[v].size(); ++j) {//找v的同性朋友 D 47 int d = adj[v][j]; 48 if(v == c || u == d) continue; 49 if(G[c][d]) ans.push_back({abs(stoi(IntToID[c])),abs(stoi(IntToID[d]))}); 50 } 51 } 52 sort(ans.begin(),ans.end(),cmp); 53 printf("%d\n",ans.size()); 54 for(int i = 0; i < ans.size(); ++i) 55 printf("%04d %04d\n",ans[i].a,ans[i].b); 56 } 57 return 0; 58 }
Method three:
Open very large array, direct mapping ID, without using a map.
1 #include<iostream> 2 #include<vector> 3 #include<algorithm> 4 5 using namespace std; 6 const int maxn = 10000; 7 struct node { 8 int a,b; 9 }; 10 int n,m,k,G[maxn][maxn]= {0}; 11 vector<int> adj[maxn]; 12 bool cmp(node a,node b) { 13 if(a.a != b.a) return a.a < b.a; 14 the else return ab & < BB; 15 } 16 int main () { . 17 CIN >> >> n- m; 18 is String A, B; . 19 for ( int I = 0 ; I <m; I ++ ) { 20 is CIN >> A> > B; // the reason for this is that there are error-prone reading point: -0000 21 is int U = ABS (Stoi (a)), V = ABS (Stoi (B)); 22 is G [U] [V] = G [V] [U] = . 1 ; 23 is IF (a.size () == b.size ()) { // sex friends 24 ADJ [U] .push_back (V); 25 adj[v].push_back(u); 26 } 27 } 28 cin>>k; 29 while(k--) { 30 cin>>a>>b; 31 vector<node> ans; 32 int u = abs(stoi(a)),v = abs(stoi(b)); 33 for(int i = 0; i < adj[u].size(); i++) { 34 int c = adj[u][i]; 35 for(int j = 0; j < adj[v].size(); j++) { 36 int d = adj[v][j]; 37 if(v == c || u == d) continue; 38 if(G[c][d]) ans.push_back({c,d}); 39 } 40 } 41 sort(ans.begin(),ans.end(),cmp); 42 printf("%d\n",ans.size()); 43 for(int i = 0; i < ans.size(); i++) { 44 printf("%04d %04d\n",ans[i].a,ans[i].b);//%04d为易错点 45 } 46 } 47 return 0 ; 48 }
