给你n个长度为3的子串， 这些子串是由一个长度为n+2的串分割得来的， 求原串， 如果给出的不合法， 输出-1。
一个欧拉通路的题， 将子串的前两个字符和后两个字符看成一个点， 比如acb， 就是ac->cb。 然后建图。
1 #include <iostream> 2 #include <vector> 3 #include <cstdio> 4 #include <cstring> 5 #include <algorithm> 6 #include <cmath> 7 #include <map> 8 #include <set> 9 #include <stack> 10 #include <string> 11 #include <queue> 12 using namespace std; 13 #define pb(x) push_back(x) 14 #define ll long long 15 #define mk(x, y) make_pair(x, y) 16 #define lson l, m, rt<<1 17 #define mem(a) memset(a, 0, sizeof(a)) 18 #define rson m+1, r, rt<<1|1 19 #define mem1(a) memset(a, -1, sizeof(a)) 20 #define mem2(a) memset(a, 0x3f, sizeof(a)) 21 #define rep(i, n, a) for(int i = a; i<n; i++) 22 #define fi first 23 #define se second 24 typedef pair<int, int> pll; 25 const double PI = acos(-1.0); 26 const double eps = 1e-8; 27 const int mod = 1e9+7; 28 const int inf = 1061109567; 29 const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} }; 30 map <string, int> ma; 31 string a[400005]; 32 int cnt, inde[400005], outde[400005], vis[400005*2], num, ans[2*400005], ecnt[66000]; 33 vector <int> v[66000]; 34 void dfs(int u) { 35 while(ecnt[u]<v[u].size()) { 36 dfs(v[u][ecnt[u]++]); 37 } 38 ans[num++] = u%256; 39 } 40 int main() 41 { 42 ios::sync_with_stdio(0); 43 string s, tmp; 44 int n, pos1, pos2; 45 cin>>n; 46 int cnt = 1, start; 47 for(int i = 0; i<n; i++) { 48 cin>>s; 49 int u = 256*s[0]+s[1]; 50 int to = 256*s[1]+s[2]; 51 v[u].pb(to); 52 inde[to]++; 53 outde[u]++; 54 start = u; 55 } 56 pos1 = -1, pos2 = -1; 57 int flag = 0; 58 for(int i = 1; i<66000; i++) { 59 if(outde[i]!=inde[i]) { 60 if(outde[i]==inde[i]+1) { 61 if(pos1==-1) { 62 pos1 = i; 63 } else { 64 flag = 1; 65 } 66 } else { 67 if(pos2==-1) 68 pos2=i; 69 else 70 flag = 1; 71 } 72 } 73 } 74 if(flag||(pos1==-1&&pos2!=-1||pos2==-1&&pos1!=-1)) { 75 cout<<"NO"<<endl; 76 return 0; 77 } 78 num = 0; 79 if(pos1 == -1) 80 pos1 = start; 81 dfs(pos1); 82 if(num != n+1) { 83 puts("NO"); 84 return 0; 85 } 86 s = "", tmp = ""; 87 tmp = (char)(pos1/256); 88 for(int i = num-1; i>=0; i--) { 89 tmp += char(ans[i]); 90 } 91 cout<<"YES"<<endl; 92 cout<<tmp; 93 return 0; 94 }