hdu5348 MZL's endless loop(欧拉回路)-优快云博客

本文介绍了一种将无向图转换为有向图的方法，确保每个节点的出入度差不超过1。通过构建欧拉回路并在奇数度节点间补充边来实现。文章还分享了一个ACM竞赛题目的解决方案。

转载请注明出处： http://www.cnblogs.com/fraud/ ——by fraud

MZL's endless loop

Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1705 Accepted Submission(s): 369
Special Judge

Problem Description

As we all kown, MZL hates the endless loop deeply, and he commands you to solve this problem to end the loop.
You are given an undirected graph with

Input

The first line of the input is a single integer

Output

For each test case, if there is no solution, print a single line with

Sample Input

3 3

1 2

2 3

3 1

7 6

1 2

1 3

1 4

1 5

1 6

1 7

Sample Output

题意就是给你一张无向图，让你把它变成有向图，使得对于每一个顶点都满足出度与入度的差的绝对值小于等于一

利用欧拉回路，在欧拉图中，每个点的出度都等于入度，那么对于这个图，其实就相当于若干个欧拉图，然后去掉一些边。

然后我们需要做的就是补边。也就是对于每个奇度点，加一条连到其它奇度点的无向边，然后跑欧拉回路，跑的方向就是这条边的方向。

另外注意有多个连通分支。这题比较容易T，虽然我的队友在比赛时瞬间就AC了。。。然而我还是在赛后T了好久，毕竟队友是final选手

  1 /**
  2  * code generated by JHelper
  3  * More info: https://github.com/AlexeyDmitriev/JHelper
  4  * @author xyiyy @https://github.com/xyiyy
  5  */
  6 
  7 #include <iostream>
  8 #include <fstream>
  9 
 10 //#####################
 11 //Author:fraud
 12 //Blog: http://www.cnblogs.com/fraud/
 13 //#####################
 14 #pragma comment(linker, "/STACK:102400000,102400000")
 15 #include <iostream>
 16 #include <sstream>
 17 #include <ios>
 18 #include <iomanip>
 19 #include <functional>
 20 #include <algorithm>
 21 #include <vector>
 22 #include <string>
 23 #include <list>
 24 #include <queue>
 25 #include <deque>
 26 #include <stack>
 27 #include <set>
 28 #include <map>
 29 #include <cstdio>
 30 #include <cstdlib>
 31 #include <cmath>
 32 #include <cstring>
 33 #include <climits>
 34 #include <cctype>
 35 
 36 using namespace std;
 37 #define rep(X, N) for(int X=0;X<N;X++)
 38 
 39 const int MAXN = 800010;
 40 int head[MAXN];
 41 int Next[MAXN], To[MAXN];
 42 int vis[MAXN];
 43 int used[100010];
 44 int deg[100010];
 45 int gao;
 46 int tot;
 47 
 48 void init(int n) {
 49     tot = 0;
 50     rep(i, n)head[i] = -1;
 51 }
 52 
 53 void addedge(int u, int v) {
 54     Next[tot] = head[u];
 55     To[tot] = v;
 56     vis[tot] = 0;
 57     head[u] = tot++;
 58 }
 59 
 60 void eular(int u){
 61     used[u] = 1;
 62     int i;
 63     while(head[u]!=-1){
 64     //for(int &i = head[u];i != -1;i = Next[i]){
 65         i = head[u];
 66         head[u] = Next[head[u]];
 67         if(vis[i])continue;
 68         vis[i^1] = 1;
 69         eular(To[i]);
 70     }
 71 }
 72 int Scan() {
 73     int res=0, ch;
 74     while(ch=getchar(), ch<'0'||ch>'9');
 75     res=ch-'0';
 76     while((ch=getchar())>='0'&&ch<='9')
 77         res=res*10+ch-'0';
 78     return res;
 79 }
 80 void Out(int a) {
 81     if (a > 9)
 82         Out(a / 10);
 83     putchar(a % 10 + '0');
 84 }
 85 
 86 class hdu5348 {
 87 public:
 88     void solve() {
 89         int t;
 90         t =Scan();//in >> t;
 91         while (t--) {
 92             int n, m;
 93             n = Scan();m=Scan();//in >> n >> m;
 94             init(n);
 95             rep(i, n)deg[i] = 0;
 96             int u, v;
 97             rep(i, m) {
 98                 u = Scan();v= Scan();//in >> u >> v;
 99                 u--, v--;
100                 deg[u]++;
101                 deg[v]++;
102                 addedge(u, v);
103                 addedge(v, u);
104             }
105             gao = -1;
106             rep(i, n) {
107                 if (deg[i] & 1) {
108                     if (gao != -1) {
109                         addedge(i, gao);
110                         addedge(gao, i);
111                         gao = -1;
112                     } else gao = i;
113                 }
114             }
115             rep(i, n) used[i] = 0;
116             /*rep(i,n){
117                 if(!used[i]){
118                     dfs(i);
119                     num++;
120                 }
121             }*/
122             gao = -1;
123             rep(i, n) {
124                 if (!used[i]) {
125                     eular(i);
126                 }
127             }
128             m<<=1;
129             for(int i=1;i<m;i+=2){
130                 if (vis[i])putchar('1');
131                 else putchar('0');
132                 putchar('\n');
133             }
134 
135         }
136     }
137 };
138 
139 
140 int main() {
141     //std::ios::sync_with_stdio(false);
142     //std::cin.tie(0);
143     hdu5348 solver;
144     //std::istream &in(std::cin);
145     //std::ostream &out(std::cout);
146     solver.solve();
147     return 0;
148 }