Sightseeing tour(混合图欧拉回路判定)

时间：2021-03-08 16:16:38 阅读：33 评论：0 收藏：0 [点我收藏+]

The city executive board in Lund wants to construct a sightseeing tour by bus in Lund, so that tourists can see every corner of the beautiful city. They want to construct the tour so that every street in the city is visited exactly once. The bus should also start and end at the same junction. As in any city, the streets are either one-way or two-way, traffic rules that must be obeyed by the tour bus. Help the executive board and determine if it‘s possible to construct a sightseeing tour under these constraints.

Input

On the first line of the input is a single positive integer n, telling the number of test scenarios to follow. Each scenario begins with a line containing two positive integers m and s, 1 <= m <= 200,1 <= s <= 1000 being the number of junctions and streets, respectively. The following s lines contain the streets. Each street is described with three integers, xi, yi, and di, 1 <= xi,yi <= m, 0 <= di <= 1, where xi and yi are the junctions connected by a street. If di=1, then the street is a one-way street (going from xi to yi), otherwise it‘s a two-way street. You may assume that there exists a junction from where all other junctions can be reached.

Output

For each scenario, output one line containing the text "possible" or "impossible", whether or not it‘s possible to construct a sightseeing tour.

Sample Input

Sample Output

possible
impossible
impossible
possible

Sponsor

题目的意思是给出一张图，有有向边和无向边，稳是否存在欧拉回路
思路：

混合图的欧拉回路判定，

先把无向边随意值一个方向，判断每个点入度出度之差，若为奇数不可能。否则进行进一步判定。

你把无向边安指定的方向建图，算出每个点的入度出度，

若入度大于出度则吧源点和它相连，反之连汇点，判断是否满流

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <string>
  5 #include <algorithm>
  6 #include <cmath>
  7 #include <map>
  8 #include <set>
  9 #include <stack>
 10 #include <queue>
 11 #include <vector>
 12 #include <bitset>
 13  
 14 using namespace std;
 15  
 16 #define LL long long
 17 const int INF = 0x3f3f3f3f;
 18 #define MAXN 500
 19  
 20 struct node
 21 {
 22     int u, v, next, cap;
 23 } edge[MAXN*MAXN];
 24 int nt[MAXN], s[MAXN], d[MAXN], visit[MAXN];
 25 int cnt;
 26 int a[2005];
 27 int b[2005];
 28 struct ed
 29 {
 30     int u,v,f;
 31 } e[2005];
 32  
 33 void init()
 34 {
 35     cnt = 0;
 36     memset(s, -1, sizeof(s));
 37 }
 38  
 39 void add(int u, int v, int c)
 40 {
 41     edge[cnt].u = u;
 42     edge[cnt].v = v;
 43     edge[cnt].cap = c;
 44     edge[cnt].next = s[u];
 45     s[u] = cnt++;
 46     edge[cnt].u = v;
 47     edge[cnt].v = u;
 48     edge[cnt].cap = 0;
 49     edge[cnt].next = s[v];
 50     s[v] = cnt++;
 51 }
 52  
 53 bool BFS(int ss, int ee)
 54 {
 55     memset(d, 0, sizeof d);
 56     d[ss] = 1;
 57     queue<int>q;
 58     q.push(ss);
 59     while (!q.empty())
 60     {
 61         int pre = q.front();
 62         q.pop();
 63         for (int i = s[pre]; ~i; i = edge[i].next)
 64         {
 65             int v = edge[i].v;
 66             if (edge[i].cap > 0 && !d[v])
 67             {
 68                 d[v] = d[pre] + 1;
 69                 q.push(v);
 70             }
 71         }
 72     }
 73     return d[ee];
 74 }
 75  
 76 int DFS(int x, int exp, int ee)
 77 {
 78     if (x == ee||!exp) return exp;
 79     int temp,flow=0;
 80     for (int i = nt[x]; ~i ; i = edge[i].next, nt[x] = i)
 81     {
 82         int v = edge[i].v;
 83         if (d[v] == d[x] + 1&&(temp = (DFS(v, min(exp, edge[i].cap), ee))) > 0)
 84         {
 85             edge[i].cap -= temp;
 86             edge[i ^ 1].cap += temp;
 87             flow += temp;
 88             exp -= temp;
 89             if (!exp) break;
 90         }
 91     }
 92     if (!flow) d[x] = 0;
 93     return flow;
 94 }
 95  
 96 int Dinic_flow(int ss, int ee)
 97 {
 98     int ans = 0;
 99     while (BFS(ss, ee))
100     {
101         for (int i = 0; i <= ee; i++) nt[i] = s[i];
102         ans+= DFS(ss, INF, ee);
103     }
104     return ans;
105 }
106  
107 int main()
108 {
109     int T;
110     int n,m;
111     for(scanf("%d",&T); T--;)
112     {
113         init();
114         scanf("%d%d",&n,&m);
115         memset(a,0,sizeof a);
116         memset(b,0,sizeof b);
117         for(int i=0; i<m; i++)
118         {
119             scanf("%d%d%d",&e[i].u,&e[i].v,&e[i].f);
120             a[e[i].u]++,b[e[i].v]++;
121         }
122         int flag=0;
123         for(int i=1; i<=n; i++)
124             if(((int)fabs(a[i]-b[i]))%2==1)
125             {
126                 flag=1;
127                 break;
128             }
129         if(!flag)
130         {
131             memset(a,0,sizeof a);
132             memset(b,0,sizeof b);
133             for(int i=0; i<m; i++)
134             {
135                 if(e[i].f==0)
136                     add(e[i].u,e[i].v,1);
137  
138                     a[e[i].u]++,b[e[i].v]++;
139             }
140             int ct=0;
141             for(int i=1; i<=n; i++)
142                 if(a[i]<b[i])
143                     add(i,n+1,(b[i]-a[i])/2);
144                 else if(a[i]>b[i])
145                     add(0,i,(a[i]-b[i])/2),ct+=(a[i]-b[i])/2;
146             if(ct!=Dinic_flow(0,n+1))
147             flag=1;
148         }
149         if(flag)
150             printf("impossible\n");
151         else
152             printf("possible\n");
153     }
154     return 0;
155 }

Sightseeing tour(混合图欧拉回路判定)

原文：https://www.cnblogs.com/csx-zzh/p/14499255.html

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