HDU3018 几笔画(非1笔)
Ant Trip
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2010 Accepted Submission(s): 782
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<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn=;
int father[maxn];
int indegree[maxn];
int n,m;
int a[maxn];
void init(){
for(int i=;i<=n;i++)
father[i]=i;
}
int find(int u){
if(father[u]!=u)
father[u]=find(father[u]);
return father[u];
}
void Union(int x,int y){
int t1=find(x);
int t2=find(y);
if(t1!=t2)
father[t1]=t2;
}
int main(){
while(scanf("%d%d",&n,&m)!=EOF){
init();
// memset(father,0,sizeof(father));
memset(indegree,,sizeof(indegree));
memset(a,-,sizeof(a));
for(int i=;i<=m;i++){
int u,v;
scanf("%d%d",&u,&v);
indegree[u]++;
indegree[v]++;
Union(u,v);
} for(int i=;i<=n;i++){
int t=find(i);
if(a[t]==-&&indegree[i]>)
a[t]=;
if(indegree[i]&)
a[t]++;
}
int ans=;
for(int i=;i<=n;i++){
if(a[i]==)
ans++;
else if(a[i]>)
ans+=a[i]/;
}
printf("%d\n",ans);
}
return ;
}
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