Euler Circuit UVA - 10735(混合图输出路径)

就是求混合图是否存在欧拉回路

如果存在则输出一组路径

 

(我就说嘛 咱的代码怎么可能错。。。。。最后的输出格式竟然w了一天 我都没发现)

解析:

  对于无向边定向建边放到网络流图中add(u, v, 1);

  对于有向边放到另一个图中add2(u, v);

  然后就是混合边求是否有欧拉  

  一边dinic后 遍历每一条边 如果不是反向边 且 起点不是s 终点不是t 

  如果Node[i].c == 0 则 add2(Node[i].v, Node[i].u);

  else add2(Node[i].u, Node[i].v);

  然后用有向图的fleury输出边就好了

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <sstream>
  4 #include <cstring>
  5 #include <map>
  6 #include <cctype>
  7 #include <set>
  8 #include <vector>
  9 #include <stack>
 10 #include <queue>
 11 #include <algorithm>
 12 #include <cmath>
 13 #include <bitset>
 14 #define rap(i, a, n) for(int i=a; i<=n; i++)
 15 #define rep(i, a, n) for(int i=a; i<n; i++)
 16 #define lap(i, a, n) for(int i=n; i>=a; i--)
 17 #define lep(i, a, n) for(int i=n; i>a; i--)
 18 #define rd(a) scanf("%d", &a)
 19 #define rlld(a) scanf("%lld", &a)
 20 #define rc(a) scanf("%c", &a)
 21 #define rs(a) scanf("%s", a)
 22 #define pd(a) printf("%d\n", a);
 23 #define plld(a) printf("%lld\n", a);
 24 #define pc(a) printf("%c\n", a);
 25 #define ps(a) printf("%s\n", a);
 26 #define MOD 2018
 27 #define LL long long
 28 #define ULL unsigned long long
 29 #define Pair pair<int, int>
 30 #define mem(a, b) memset(a, b, sizeof(a))
 31 #define _  ios_base::sync_with_stdio(0),cin.tie(0)
 32 //freopen("1.txt", "r", stdin);
 33 using namespace std;
 34 const int maxn = 100010, INF = 2e9;
 35 int n, m, s, t, cnt;
 36 int in[maxn], out[maxn], vis[maxn];
 37 int d[maxn], head[maxn], cur[maxn];
 38 set<int> ss;
 39 int st[maxn],cnt3;
 40 int cnt2, head2[maxn];
 41 
 42 
 43 struct edge
 44 {
 45     int u, v, c, next, ff;
 46 }Edge[maxn << 1];
 47 
 48 void add_(int u, int v, int c, int ff)
 49 {
 50     Edge[cnt].u = u;
 51     Edge[cnt].v = v;
 52     Edge[cnt].c = c;
 53     Edge[cnt].ff = ff;
 54     Edge[cnt].next = head[u];
 55     head[u] = cnt++;
 56 }
 57 
 58 void add(int u, int v, int c)
 59 {
 60     add_(u, v, c, 1);
 61     add_(v, u, 0, 0);
 62 }
 63 
 64 bool bfs()
 65 {
 66     queue<int> Q;
 67     mem(d, 0);
 68     Q.push(s);
 69     d[s] = 1;
 70     while(!Q.empty())
 71     {
 72         int u = Q.front(); Q.pop();
 73         for(int i = head[u]; i != -1; i = Edge[i].next)
 74         {
 75             edge e = Edge[i];
 76             if(!d[e.v] && e.c > 0)
 77             {
 78                 d[e.v] = d[e.u] + 1;
 79                 Q.push(e.v);
 80                 if(e.v == t) return 1;
 81             }
 82         }
 83     }
 84     return d[t] != 0;
 85 }
 86 
 87 int dfs(int u, int cap)
 88 {
 89     int ret = 0;
 90     if(u == t || cap == 0)
 91         return cap;
 92     for(int &i = cur[u]; i != -1; i = Edge[i].next)
 93     {
 94         edge e = Edge[i];
 95         if(d[e.v] == d[u] + 1 && e.c > 0)
 96         {
 97             int V = dfs(e.v, min(cap, e.c));
 98             Edge[i].c -= V;
 99             Edge[i^1].c += V;
100             ret += V;
101             cap -= V;
102             if(cap == 0) break;
103         }
104     }
105     if(cap > 0) d[u] = -1;
106     return ret;
107 }
108 
109 int Dinic(int u)
110 {
111     int ans = 0;
112     while(bfs())
113     {
114         memcpy(cur, head, sizeof(head));
115         ans += dfs(u, INF);
116     }
117     return ans;
118 }
119 
120 
121 struct node
122 {
123     int u, v, flag, next;
124 }Node[maxn << 1];
125 
126 void add2(int u, int v)
127 {
128     Node[cnt2].u = u;
129     Node[cnt2].v = v;
130     Node[cnt2].next = head2[u];
131     Node[cnt2].flag = 0;
132     head2[u] = cnt2++;
133 }
134 int used[maxn];
135 void fleury(int u)
136 {
137     for(int i = head2[u]; i != -1; i = Node[i].next)
138     {
139         if(!used[i])
140         {
141             used[i] = 1;
142             fleury(Node[i].v);
143         }
144 
145     }
146     st[cnt3++] = u;
147 }
148 
149 
150 void init()
151 {
152     mem(in, 0);
153     mem(head, -1);
154     mem(out, 0);
155     mem(st, 0);
156     cnt = 0;
157     cnt2 = 0;
158     cnt3 = 0;
159     mem(head2, -1);
160     mem(used, 0);
161     ss.clear();
162     
163 }
164 
165 char str[2];
166 
167 int main()
168 {
169     int T;
170     cin >> T;
171     while(T--)
172     {
173         int u, v, w;
174         cin >> n >> m;
175         init();
176         s = 0, t = maxn - 1;
177         for(int i = 1; i <= m; i++)
178         {
179             scanf("%d%d%s", &u, &v, str);
180             in[v]++, out[u]++;
181             if(str[0] == 'U') add(u, v, 1);
182             else if(str[0] == 'D') add2(u, v);
183         }
184         int flag = 0, m_sum = 0;
185         for(int i = 1; i <= n; i++)
186         {
187             if(abs(out[i] - in[i]) & 1)
188             {
189                 flag = 1;
190                 break;
191             }
192             if(out[i] > in[i]) add(s, i, (out[i] - in[i]) / 2), m_sum += (out[i] - in[i]) / 2;
193             else if(in[i] > out[i]) add(i, t, (in[i] - out[i]) / 2);
194 
195         }
196         if(!flag && m_sum == Dinic(s))
197         {
198             for(int i = 0; i < cnt; i++)
199             {
200                 if(!Edge[i].ff || Edge[i].u == s || Edge[i].v == t || Edge[i].u == t || Edge[i].v == s) continue;
201                 if(Edge[i].c == 0) add2(Edge[i].v, Edge[i].u);
202                 else add2(Edge[i].u, Edge[i].v);
203             }
204             fleury(1);
205             for(int i = cnt3 - 1; i >= 0; i--)
206             {
207                 if(i != cnt3 - 1) printf(" ");
208                 printf("%d", st[i]);
209             }
210 
211 
212             printf("\n");
213 
214 
215         }
216         else
217             cout << "No euler circuit exist" << endl;
218         if(T) printf("\n");
219 
220     }
221 
222     return 0;
223 }

 

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