hdu 3018 Ant Trip （欧拉图+并查集）

Ant Trip

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2727    Accepted Submission(s): 1075

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 31 22 31 34 21 23 4

 

Sample Output

12
Hint
New ~~~ Notice: if there are no road connecting one town ,tony may forget about the town.In sample 1,tony and his friends just form one group,they can start at either town 1,2,or 3.In sample 2,tony and his friends must form two group.
 

 

Source

2009 Multi-University Training Contest 12 - Host by FZU

 

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题解：欧拉图

用并查集维护连通性。对于同一个连通块中的点，如果所有点的度都是偶数，那么该连通块中存在欧拉回路。如果有K个奇数度的结点，那么需要k/2条路径来覆盖。

#include<iostream>#include<cstring>#include<algorithm>#include<cstdio>#include<cmath>#define N 500003using namespace std;int n,m,tot,nxt[N],point[N],v[N],cnt,sz,x[N],y[N],ans,fa[N];int dfsn[N],low[N],st[N],top,ins[N],belong[N],du[N],size[N];struct data{int bl,x;}a[N];int cmp(data a,data b){return a.bl<b.bl;}int find(int x){if (fa[x]==x) return x;fa[x]=find(fa[x]);return fa[x];}int main(){freopen("a.in","r",stdin);freopen("my.out","w",stdout);while (scanf("%d%d",&n,&m)!=EOF) {memset(du,0,sizeof(du));tot=0; ans=0;for (int i=1;i<=n;i++) fa[i]=i;for (int i=1;i<=m;i++) {    scanf("%d%d",&x[i],&y[i]);int r1=find(x[i]); int r2=find(y[i]);if (r2!=r1) fa[r2]=r1;du[x[i]]++; du[y[i]]++;}for (int i=1;i<=n;i++) fa[i]=find(i),a[i].x=i,a[i].bl=fa[i];sort(a+1,a+n+1,cmp);int k=0; int s=0;//for (int i=1;i<=n;i++) cout<<du[i]<<" ";//cout<<endl;for (int i=1;i<=n;i++)  if (a[i].bl!=a[i-1].bl||i==n) { if (i==n&&a[i].bl==a[i-1].bl) {             s++; if (du[a[i].x]&1) k++;     }    //cout<<k<<" "<<s<<" "<<ans<<endl; if (k) ans+=k/2; else if (s>1) ans++; k=0; s=1; if (du[a[i].x]&1) k=1; }else {   if (du[a[i].x]&1) k++;   s++; }printf("%d\n",ans);}}