HDU 6311 最少路径覆盖边集 欧拉路径

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Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1543    Accepted Submission(s): 321
Special Judge

Problem Description

The Wall has down and the King in the north has to send his soldiers to sentinel.
The North can be regard as a undirected graph (not necessary to be connected), one soldier can cover one path. Today there's no so many people still breathing in the north, so the King wants to minimize the number of soldiers he sent to cover each edge exactly once. As a master of his, you should tell him how to arrange soldiers.

 

Input

There might be multiple test cases, no more than 20. You need to read till the end of input.
In the first line, two integers n and m, representing the number of nodes and edges in the graph.
In the following m lines, each contain two integers, representing two ends of an edge.
There are no parallel edges or self loops.
1≤n,m≤100000

 

Output

For each test case, the first line contains number of needed routes, p.
For the following p lines, an integer x in the beginning, followed by x integers, representing the list of used edges. Every integer should be a positive or negative integer. Its absolute value represents the number of chosen edge (1~n). If it's positive, it shows that this edge should be passed as the direction as the input, otherwise this edge should be passed in the direction different from the input. Edges should be in correct order.

 

Sample Input

3 3

1 2

1 3

2 3

 

Sample Output

1

3 1 3 -2

 

Source

2018 Multi-University Training Contest 2

解析  对于每个连通块最少路径覆盖就是  max(1,度数为奇数点的个数/2)  然后我们把每两个奇数点连一条虚边 要剩两个 我们求欧拉路径或者欧拉回路 都可以求解  求得的路径删掉虚边 剩下的线段就是最少的路径  但是 求欧拉路径比较容易计算路径条数 所以剩两个从一个奇数点出发回到另一个奇数点   （没有奇数点就从任意一点出发 得到欧拉回路 没有虚边 一条路就可以走完 ） 独立点不要算进来 覆盖所有的边 不是所有的点。

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欧拉回路 或者 欧拉路径都可以直接dfs得到  回朔的时候保存路径就好了 因为走死一个圈 退回去 转一圈还可以回来

欧拉回路求解算法：

https://blog.csdn.net/u011466175/article/details/18861415

AC代码

 #include<bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 const int maxn=1e5+,mod=1e9+;
 const ll inf=1e16;
 #define pb push_back
 #define mp make_pair
 struct edge
 {
     int from,to;
     edge(int u,int v):from(u),to(v) {}
 };
 int n,m,cnt;
 vector<int> ans[maxn],ji;
 vector<edge> edges;
 vector<int> g[maxn];
 int du[maxn];
 int p[maxn*],vis[maxn];//p标记边是否用过要开大一点 因为自己还要加边，vis标记点是否访问过
 void init()
 {
     cnt=;
     for(int i=; i<=n; i++) g[i].clear(),ans[i].clear();
     memset(vis,,sizeof(vis));
     memset(du,,sizeof(du));
     memset(p,,sizeof(p));
     edges.clear();
 }
 void addedge(int from,int to)
 {
     edges.push_back(edge(from,to));
     edges.push_back(edge(to,from));
     int siz=edges.size();
     g[from].push_back(siz-);
     g[to].push_back(siz-);
 }
 void dfs(int x)
 {
     vis[x]=;
     if(du[x]%)
         ji.pb(x);
     for(int i=;i<g[x].size();i++)
     {
         edge e=edges[g[x][i]];
         if(!vis[e.to])
             dfs(e.to);
     }
 }
 void dfs2(int x)
 {
     for(int i=;i<g[x].size();i++)
     {
         int u=g[x][i];
         edge e=edges[u];
         if(!p[u])
         {
             p[u]=p[u^]=;
             dfs2(e.to);
             int temp=u%?-u/-:u/+;
             if(u<*m)
                 ans[cnt].pb(temp);
             else
                 cnt++;
         }
     }
 }
 int main()
 {
     while(~scanf("%d%d",&n,&m))
     {
         int u,v;init();
         for(int i=;i<=m;i++)
         {
             scanf("%d%d",&u,&v);
             addedge(u,v);
             du[u]++,du[v]++;
         }
         for(int i=;i<=n;i++)
         {
             if(!vis[i]&&du[i])
             {
                 ji.clear();
                 dfs(i);
                 for(int i=;i<ji.size();i+=)
                 {
                     addedge(ji[i],ji[i+]);
                 }
                 int t=ji.size()?ji[]:i;
                 dfs2(t);
                 cnt++;
             }
         }
         printf("%d\n",cnt);
         for(int i=;i<cnt;i++)
         {
             printf("%d",ans[i].size());
             for(int j=ans[i].size()-;j>=;j--)
             {
                 printf(" %d",ans[i][j]);
             }
             printf("\n");
         }
     }
 }