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 << "]" << "
";
    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, 1, BUF, stdin);
    return *bg++;
} 

inline int ri(){
    int x = 0, f = 1, c = gc();
    for(; c<48||c>57; f = c=='-'?-1:f, c=gc());
    for(; c>47&&c<58; x = x*10 + c - 48, 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-10;
const LL inf = 0x7fffffff;
const LL infLL = 0x7fffffffffffffffLL;
const LL mod = 1e9 + 7;
const int maxN = 1e5 + 7;
const LL ONE = 1;
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 = 0;
    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] = 1;
        hierholzer(t.to);
        a[cnt--] = t.to;
    }
}

int main(){
    INIT();
    cin >> n;
    For(i, 1, n - 1) cin >> b[i];
    For(i, 1, n - 1) {
        cin >> c[i];
        if(c[i] < b[i]) {
            cout << -1 << endl;
            return 0;
        }
    }
    For(i, 1, n - 1) {
        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 % 2 == 1) {
            ++cnt;
            S = i->ft;
        }
    }
    if(S == 0) S = b[1]; // 没有奇度顶点就随便选一个顶点 
    if(cnt > 2 || cnt == 1) { // 有两个或0个奇度顶点，才可能有欧拉通路 
        cout << -1 << endl;
        return 0;
    }
    
    cnt = n;
    hierholzer(S);
    a[cnt--] = S;
    
    // 整个图可能不连通 
    if(cnt == 0)For(i, 1, n) cout << a[i] << " ";
    else cout << -1;
    cout << endl;
    return 0;
}