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

  1 #include <bits/stdc++.h>
  2 using namespace std;
  3  
  4 #define INIT() std::ios::sync_with_stdio(false);std::cin.tie(0);
  5 #define Rep(i,n) for (int i = 0; i < (n); ++i)
  6 #define For(i,s,t) for (int i = (s); i <= (t); ++i)
  7 #define rFor(i,t,s) for (int i = (t); i >= (s); --i)
  8 #define ForLL(i, s, t) for (LL i = LL(s); i <= LL(t); ++i)
  9 #define rForLL(i, t, s) for (LL i = LL(t); i >= LL(s); --i)
 10 #define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)
 11 #define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i)
 12  
 13 #define pr(x) cout << #x << " = " << x << "  "
 14 #define prln(x) cout << #x << " = " << x << endl
 15  
 16 #define LOWBIT(x) ((x)&(-x))
 17  
 18 #define ALL(x) x.begin(),x.end()
 19 #define INS(x) inserter(x,x.begin())
 20  
 21 #define ms0(a) memset(a,0,sizeof(a))
 22 #define msI(a) memset(a,inf,sizeof(a))
 23 #define msM(a) memset(a,-1,sizeof(a))
 24 
 25 #define MP make_pair
 26 #define PB push_back
 27 #define ft first
 28 #define sd second
 29  
 30 template<typename T1, typename T2>
 31 istream &operator>>(istream &in, pair<T1, T2> &p) {
 32     in >> p.first >> p.second;
 33     return in;
 34 }
 35  
 36 template<typename T>
 37 istream &operator>>(istream &in, vector<T> &v) {
 38     for (auto &x: v)
 39         in >> x;
 40     return in;
 41 }
 42  
 43 template<typename T1, typename T2>
 44 ostream &operator<<(ostream &out, const std::pair<T1, T2> &p) {
 45     out << "[" << p.first << ", " << p.second << "]" << "\n";
 46     return out;
 47 }
 48 
 49 inline int gc(){
 50     static const int BUF = 1e7;
 51     static char buf[BUF], *bg = buf + BUF, *ed = bg;
 52     
 53     if(bg == ed) fread(bg = buf, 1, BUF, stdin);
 54     return *bg++;
 55 } 
 56 
 57 inline int ri(){
 58     int x = 0, f = 1, c = gc();
 59     for(; c<48||c>57; f = c=='-'?-1:f, c=gc());
 60     for(; c>47&&c<58; x = x*10 + c - 48, c=gc());
 61     return x*f;
 62 }
 63  
 64 typedef long long LL;
 65 typedef unsigned long long uLL;
 66 typedef pair< double, double > PDD;
 67 typedef pair< int, int > PII;
 68 typedef pair< string, int > PSI;
 69 typedef set< int > SI;
 70 typedef vector< int > VI;
 71 typedef map< int, int > MII;
 72 typedef pair< LL, LL > PLL;
 73 typedef vector< LL > VL;
 74 typedef vector< VL > VVL;
 75 const double EPS = 1e-10;
 76 const LL inf = 0x7fffffff;
 77 const LL infLL = 0x7fffffffffffffffLL;
 78 const LL mod = 1e9 + 7;
 79 const int maxN = 1e5 + 7;
 80 const LL ONE = 1;
 81 const LL evenBits = 0xaaaaaaaaaaaaaaaa;
 82 const LL oddBits = 0x5555555555555555;
 83 
 84 struct Edge{
 85     int id;
 86     int from, to;
 87     
 88     Edge() {}
 89     Edge(int id, int from, int to) : id(id), from(from), to(to) {}
 90 };
 91 
 92 struct Vertex{
 93     int in = 0;
 94     vector< Edge > edges;
 95 };
 96 
 97 int n;
 98 int b[maxN], c[maxN], a[maxN];
 99 unordered_map< int, Vertex > mIV;
100 int vis[maxN]; 
101 int S;
102 int cnt;
103 
104 //  Hierholzer算法求欧拉路 
105 inline void hierholzer(int s) {
106     vector< Edge > &tmp = mIV[s].edges;
107     while(!tmp.empty()) {
108         Edge t = tmp.back();
109         tmp.pop_back();
110         
111         if(vis[t.id]) continue;
112         vis[t.id] = 1;
113         hierholzer(t.to);
114         a[cnt--] = t.to;
115     }
116 }
117 
118 int main(){
119     INIT();
120     cin >> n;
121     For(i, 1, n - 1) cin >> b[i];
122     For(i, 1, n - 1) {
123         cin >> c[i];
124         if(c[i] < b[i]) {
125             cout << -1 << endl;
126             return 0;
127         }
128     }
129     For(i, 1, n - 1) {
130         mIV[b[i]].edges.PB(Edge(i, b[i], c[i]));
131         ++mIV[b[i]].in;
132         mIV[c[i]].edges.PB(Edge(i, c[i], b[i]));
133         ++mIV[c[i]].in;
134     }
135     
136     // 找起点
137     // cnt记录奇度顶点个数 
138     foreach(i, mIV) {
139         if(i->sd.in % 2 == 1) {
140             ++cnt;
141             S = i->ft;
142         }
143     }
144     if(S == 0) S = b[1]; // 没有奇度顶点就随便选一个顶点 
145     if(cnt > 2 || cnt == 1) { // 有两个或0个奇度顶点，才可能有欧拉通路 
146         cout << -1 << endl;
147         return 0;
148     }
149     
150     cnt = n;
151     hierholzer(S);
152     a[cnt--] = S;
153     
154     // 整个图可能不连通 
155     if(cnt == 0)For(i, 1, n) cout << a[i] << " ";
156     else cout << -1;
157     cout << endl;
158     return 0;
159 }

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