CodeForces 1152E Neko and Flashback

题目链接：http://codeforces.com/problemset/problem/1152/E

题目大意

　　有一个 1~n-1 的排列p 和长度为 n 的数组 a，数组b，c定义如下：

　　　　b：b_i = min(a_i, a_{i + 1})，1 <= i <= n-1。

　　　　c：c_i = max(a_i, a_{i + 1})，1 <= i <= n-1。

　　数组b'，c'定义如下：

　　　　b'：b'_i = b_pi，1 <= i <= n-1。

　　　　c'：c'_i = c_pi，1 <= i <= n-1。

　　给定数组 b'，c'，求 a。

分析

　　一笔画问题，求无向图欧拉通路。

　　这里用了Hierholzer算法（DFS）。

代码如下

 #include <bits/stdc++.h>
 using namespace std;

 #define INIT() std::ios::sync_with_stdio(false);std::cin.tie(0);
 #define Rep(i,n) for (int i = 0; i < (n); ++i)
 #define For(i,s,t) for (int i = (s); i <= (t); ++i)
 #define rFor(i,t,s) for (int i = (t); i >= (s); --i)
 #define ForLL(i, s, t) for (LL i = LL(s); i <= LL(t); ++i)
 #define rForLL(i, t, s) for (LL i = LL(t); i >= LL(s); --i)
 #define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)
 #define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i)

 #define pr(x) cout << #x << " = " << x << "  "
 #define prln(x) cout << #x << " = " << x << endl

 #define LOWBIT(x) ((x)&(-x))

 #define ALL(x) x.begin(),x.end()
 #define INS(x) inserter(x,x.begin())

 #define ms0(a) memset(a,0,sizeof(a))
 #define msI(a) memset(a,inf,sizeof(a))
 #define msM(a) memset(a,-1,sizeof(a))

 #define MP make_pair
 #define PB push_back
 #define ft first
 #define sd second

 template<typename T1, typename T2>
 istream &operator>>(istream &in, pair<T1, T2> &p) {
     in >> p.first >> p.second;
     return in;
 }

 template<typename T>
 istream &operator>>(istream &in, vector<T> &v) {
     for (auto &x: v)
         in >> x;
     return in;
 }

 template<typename T1, typename T2>
 ostream &operator<<(ostream &out, const std::pair<T1, T2> &p) {
     out << "[" << p.first << ", " << p.second << "]" << "\n";
     return out;
 }

 inline int gc(){
     static const int BUF = 1e7;
     static char buf[BUF], *bg = buf + BUF, *ed = bg;

     if(bg == ed) fread(bg = buf, , BUF, stdin);
     return *bg++;
 } 

 inline int ri(){
     int x = , f = , c = gc();
     for(; c<||c>; f = c=='-'?-:f, c=gc());
     for(; c>&&c<; x = x* + c - , c=gc());
     return x*f;
 }

 typedef long long LL;
 typedef unsigned long long uLL;
 typedef pair< double, double > PDD;
 typedef pair< int, int > PII;
 typedef pair< string, int > PSI;
 typedef set< int > SI;
 typedef vector< int > VI;
 typedef map< int, int > MII;
 typedef pair< LL, LL > PLL;
 typedef vector< LL > VL;
 typedef vector< VL > VVL;
 const double EPS = 1e-;
 const LL inf = 0x7fffffff;
 const LL infLL = 0x7fffffffffffffffLL;
 const LL mod = 1e9 + ;
 const int maxN = 1e5 + ;
 const LL ONE = ;
 const LL evenBits = 0xaaaaaaaaaaaaaaaa;
 const LL oddBits = 0x5555555555555555;

 struct Edge{
     int id;
     int from, to;

     Edge() {}
     Edge(int id, int from, int to) : id(id), from(from), to(to) {}
 };

 struct Vertex{
     int in = ;
     vector< Edge > edges;
 };

 int n;
 int b[maxN], c[maxN], a[maxN];
 unordered_map< int, Vertex > mIV;
 int vis[maxN];
 int S;
 int cnt;

 //  Hierholzer算法求欧拉路
 inline void hierholzer(int s) {
     vector< Edge > &tmp = mIV[s].edges;
     while(!tmp.empty()) {
         Edge t = tmp.back();
         tmp.pop_back();

         if(vis[t.id]) continue;
         vis[t.id] = ;
         hierholzer(t.to);
         a[cnt--] = t.to;
     }
 }

 int main(){
     INIT();
     cin >> n;
     For(i, , n - ) cin >> b[i];
     For(i, , n - ) {
         cin >> c[i];
         if(c[i] < b[i]) {
             cout << - << endl;
             return ;
         }
     }
     For(i, , n - ) {
         mIV[b[i]].edges.PB(Edge(i, b[i], c[i]));
         ++mIV[b[i]].in;
         mIV[c[i]].edges.PB(Edge(i, c[i], b[i]));
         ++mIV[c[i]].in;
     }

     // 找起点
     // cnt记录奇度顶点个数
     foreach(i, mIV) {
         if(i->sd.in %  == ) {
             ++cnt;
             S = i->ft;
         }
     }
     if(S == ) S = b[]; // 没有奇度顶点就随便选一个顶点
     if(cnt >  || cnt == ) { // 有两个或0个奇度顶点，才可能有欧拉通路
         cout << - << endl;
         return ;
     }

     cnt = n;
     hierholzer(S);
     a[cnt--] = S;

     // 整个图可能不连通
     if(cnt == )For(i, , n) cout << a[i] << " ";
     else cout << -;
     cout << endl;
     return ;
 }