Ant Trip
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
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.
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.
1 2
2 3
1 3
4 2
1 2
3 4
2
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.
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define mod 1000000007
#define esp 0.00000000001
const int N=1e5+,M=1e6+,inf=1e9;
const ll INF=1e18+;
int n,m;
int fa[N],du[N],flag[N],mark[N],hh[N];
int Find(int x)
{
return fa[x]==x?x:fa[x]=Find(fa[x]);
}
void update(int u,int v)
{
int x=Find(u);
int y=Find(v);
if(x!=y)
{
fa[x]=y;
}
}
void init()
{
for(int i=;i<N;i++)
fa[i]=i;
memset(flag,,sizeof(flag));
memset(du,,sizeof(du));
memset(mark,,sizeof(mark));
memset(hh,,sizeof(hh));
}
int main()
{
while(~scanf("%d%d",&n,&m))
{
init();
for(int i=;i<=m;i++)
{
int u,v;
scanf("%d%d",&u,&v);
update(u,v);
du[u]++;
du[v]++;
hh[u]=;
hh[v]=;
}
int ans=,sum=;
for(int i=;i<=n;i++)
{
if(!hh[i])continue;
int x=Find(i);
du[i]%=;
sum+=du[i];
if(!flag[x])
{
ans++;
flag[x]=;
}
if(du[i]==&&flag[x]&&!mark[x])
{
ans--;
mark[x]=;
}
}
printf("%d\n",ans+sum/);
}
return ;
}