UVa 10735 (混合图的欧拉回路) Euler Circuit

  1 #include <bits/stdc++.h>
  2 #define REP(i,n) for(int i = 0; i < (n); i++)
  3 #define PB push_back
  4 using namespace std;
  5 
  6 const int INF = 1000000000;
  7 const int maxn = 500 + 10;
  8 
  9 struct Edge
 10 {
 11     int from, to, cap, flow;
 12     Edge(int u, int v, int c, int f):from(u), to(v), cap(c), flow(f) {}
 13 };
 14 
 15 struct EdmondsKarp
 16 {
 17     int n, m;
 18     vector<Edge> edges;
 19     vector<int> G[maxn];
 20     int a[maxn], p[maxn];
 21 
 22     void Init(int n)
 23     {
 24         REP(i, n) G[i].clear();
 25         edges.clear();
 26     }
 27 
 28     void AddEdge(int from, int to, int cap)
 29     {
 30         edges.PB(Edge(from, to, cap, 0));
 31         edges.PB(Edge(to, from, 0, 0));
 32         m = edges.size();
 33         G[from].PB(m-2);
 34         G[to].PB(m-1);
 35     }
 36 
 37     int MaxFlow(int s, int t)
 38     {
 39         int flow = 0;
 40         for(;;)
 41         {
 42             queue<int> Q;
 43             Q.push(s);
 44             memset(a, 0, sizeof(a));
 45             a[s] = INF;
 46             while(!Q.empty())
 47             {
 48                 int x = Q.front(); Q.pop();
 49                 REP(i, G[x].size())
 50                 {
 51                     Edge& e = edges[G[x][i]];
 52                     if(!a[e.to] && e.cap > e.flow)
 53                     {
 54                         a[e.to] = min(a[x], e.cap - e.flow);
 55                         p[e.to] = G[x][i];
 56                         Q.push(e.to);
 57                     }
 58                 }
 59                 if(a[t]) break;
 60             }
 61             if(!a[t]) break;
 62             for(int u = t; u != s; u = edges[p[u]].from)
 63             {
 64                 edges[p[u]].flow += a[t];
 65                 edges[p[u]^1].flow -= a[t];
 66             }
 67             flow += a[t];
 68         }
 69         return flow;
 70     }
 71 }g;
 72 
 73 int n, m;
 74 int deg[maxn], u[maxn], v[maxn], id[maxn];
 75 bool directed[maxn];
 76 
 77 vector<int> G[maxn];//建新图，用来求欧拉回路
 78 vector<int> vis[maxn];
 79 vector<int> path;//欧拉回路
 80 
 81 void Euler(int u)
 82 {
 83     REP(i, G[u].size()) if(!vis[u][i])
 84     {
 85         vis[u][i] = 1;
 86         Euler(G[u][i]);
 87         path.PB(G[u][i]+1);
 88     }
 89 }
 90 
 91 void print_answer()
 92 {
 93     REP(i, n) { G[i].clear(); vis[i].clear(); }
 94     REP(i, m)
 95     {
 96         bool rev = false;
 97         if(!directed[i] && g.edges[id[i]].flow > 0) rev = true;//流量为1对应将该边反向
 98         if(!rev) { G[u[i]].PB(v[i]); vis[u[i]].PB(0); }
 99         else { G[v[i]].PB(u[i]); vis[v[i]].PB(0); }
100     }
101 
102     path.clear();
103     Euler(0);
104     printf("1");
105     for(int i = path.size()-1; i >= 0; i--) printf(" %d", path[i]);
106     puts("");
107 }
108 
109 int main()
110 {
111     //freopen("in.txt", "r", stdin);
112 
113     int T;
114     scanf("%d", &T);
115     while(T--)
116     {
117         scanf("%d%d", &n, &m);
118         g.Init(n+2);
119         memset(deg, 0, sizeof(deg));
120         REP(i, m)
121         {
122             char d;
123             scanf("%d %d %c", &u[i], &v[i], &d);
124             u[i]--; v[i]--;
125             directed[i] = (d == ‘D‘);
126             deg[u[i]]++; deg[v[i]]--;
127             if(!directed[i]) { id[i] = g.edges.size(); g.AddEdge(u[i], v[i], 1); }//第i条边在网络流中的编号
128         }
129 
130         bool ok = true;
131         REP(i, m) if(deg[i] % 2 != 0) { ok = false; break; }//出入度之和不是偶数说明不存在欧拉回路
132 
133         int s = n, t = n+1;
134         if(ok)
135         {
136             int sum = 0;
137             REP(i, n)
138             {
139                 if(deg[i] > 0) { sum += deg[i] / 2; g.AddEdge(s, i, deg[i] / 2); }
140                 if(deg[i] < 0) { g.AddEdge(i, t, -deg[i] / 2); }
141             }
142             int flow = g.MaxFlow(s, t);
143             if(flow != sum) ok = false;//最大流不满载
144         }
145 
146         if(ok) print_answer(); else puts("No euler circuit exist");
147         if(T) puts("");
148     }
149 
150     return 0;
151 }