Ant
Trip
Problem Description
Ant
Country consist of N towns.There are M roads connecting the towns.
Ant
Tony,together with his friends,wants to go through every part of the
country.
They intend to visit every road , and every road must be
visited for exact one time.However,it may be a mission impossible for only one
group of people.So they are trying to divide all the people into several
groups,and each may start at different town.Now tony wants to know what is the
least groups of ants that needs to form to achieve their goal.
Input
Input
contains multiple cases.Test cases are separated by several blank lines. Each
test case starts with two integer
N(1<=N<=100000),M(0<=M<=200000),indicating that there are N towns
and M roads in Ant Country.Followed by M lines,each line contains two integers
a,b,(1<=a,b<=N) indicating that there is a road connecting town a and town
b.No two roads will be the same,and there is no road connecting the same
town.
Output
For
each test case ,output the least groups that needs to form to achieve their
goal.
Sample Input
3 3 1 2 2 3 1 3 4 2 1 2 3 4
Sample Output
题意： 判断一共有多少个欧拉路
sl: 首先并查集维护欧拉路的个数，然后一笔画需要的次数就等与奇数度的节点+欧拉路的个数。-----并查集很神奇啊。
1 //判断有多少欧拉回路
2 #include<cstdio>
3 #include<cstring>
4 #include<algorithm>
5 #include<vector>
6 using namespace std;
7 const int MAX = 1e5+
10;
8 vector<
int> v;
9 int ans[MAX],vis[MAX];
10 int deg[MAX],odd[MAX];
11 void make(
int n)
12 {
13 for(
int i=
1;i<=n;i++) ans[i]=i;
14 }
15 int find(
int x)
16 {
17 if(ans[x]!=x)
18 ans[x]=find(ans[x]);
19 return ans[x];
20 }
21 22 void Union(
int x,
int y)
23 {
24 ans[x]=y;
25 }
26 int main()
27 {
28 int ret,n,m;
int a,b;
29 while(scanf(
"%d %d",&n,&m)==
2)
30 {
31 make(n);
32 ret=
0; memset(vis,
0,
sizeof(vis));
33 memset(odd,
0,
sizeof(odd));
34 memset(deg,
0,
sizeof(deg));
35 v.clear();
36 for(
int i=
0;i<m;i++)
37 {
38 scanf(
"%d %d",&a,&b);
39 deg[a]++; deg[b]++;
40 int x=find(a);
int y=find(b);
41 if(x!=y) Union(x,y);
42 }
43 for(
int i=
1;i<=n;i++)
44 {
45 int u=find(i); //找一个欧拉回路。
46 if(!vis[u])
47 {
48 vis[u]=
1;
49 v.push_back(u);
50 }
51 if(deg[i]&
1) odd[u]++;
52 }
53 for(
int i=
0;i<v.size();i++)
54 {
55 int k=v[i];
56 if(deg[k]==
0)
continue;
57 if(odd[k]==
0) ret++;
58 ret+=odd[k]/
2;
59 }
60 printf(
"%d\n",ret);
61 }
62 return 0;
63 }
HDU 3018 Ant Trip,布布扣,bubuko.com
HDU 3018 Ant Trip
原文：http://www.cnblogs.com/acvc/p/3643883.html
