Educational Codeforces Round 37 E. Connected Components?(920E)
2018-02-03 16:00
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题目链接:http://codeforces.com/problemset/problem/920/E
开始看成了强连通,表示不会。
其实由于边数很多,连通块的个数不会太多,用map映射一下不存在的边即可。
思路:一开始把所有点放在一个set里,每次取set中一个顶点,删去之,遍历set,删去与此顶点邻接的顶点,就得到了一个连通块,连通块大小等于删去的点的个数,反复这个过程,直到set为空。
贴代码:
#include<bits/stdc++.h>
using namespace std;
long long n,m;
set<long long> mp;
set<int> st;
bool vis[200010];
int ans[200010],ct=0;
int bfs()
{
queue<int> q;
set<int>::iterator iter=st.begin();
q.push(*st.begin());
st.erase(st.begin());
int cnt=1;
while(!q.empty())
{
long long u=q.front();
vis[u]=1;
q.pop();
for(iter=st.begin();iter!=st.end();)
{
if(!mp.count(*iter+u*n))
{
cnt++;
q.push(*iter);
st.erase(iter++);
}
else
iter++;
}
}
return cnt;
}
int main()
{
cin>>n>>m;
for(int i=0;i<m;i++)
{
long long a,b;
scanf("%I64d %I64d",&a,&b);
mp.insert(a*n+b);
mp.insert(a+b*n);
}
for(int i=1;i<=n;i++)
st.insert(i);
memset(vis,0,sizeof(vis));
while(!st.empty())
{
ans[ct]=bfs();
ct++;
}
sort(ans,ans+ct);
printf("%d\n",ct);
for(int i=0;i<ct;i++)
printf("%d ",ans[i]);
}
开始看成了强连通,表示不会。
其实由于边数很多,连通块的个数不会太多,用map映射一下不存在的边即可。
思路:一开始把所有点放在一个set里,每次取set中一个顶点,删去之,遍历set,删去与此顶点邻接的顶点,就得到了一个连通块,连通块大小等于删去的点的个数,反复这个过程,直到set为空。
贴代码:
#include<bits/stdc++.h>
using namespace std;
long long n,m;
set<long long> mp;
set<int> st;
bool vis[200010];
int ans[200010],ct=0;
int bfs()
{
queue<int> q;
set<int>::iterator iter=st.begin();
q.push(*st.begin());
st.erase(st.begin());
int cnt=1;
while(!q.empty())
{
long long u=q.front();
vis[u]=1;
q.pop();
for(iter=st.begin();iter!=st.end();)
{
if(!mp.count(*iter+u*n))
{
cnt++;
q.push(*iter);
st.erase(iter++);
}
else
iter++;
}
}
return cnt;
}
int main()
{
cin>>n>>m;
for(int i=0;i<m;i++)
{
long long a,b;
scanf("%I64d %I64d",&a,&b);
mp.insert(a*n+b);
mp.insert(a+b*n);
}
for(int i=1;i<=n;i++)
st.insert(i);
memset(vis,0,sizeof(vis));
while(!st.empty())
{
ans[ct]=bfs();
ct++;
}
sort(ans,ans+ct);
printf("%d\n",ct);
for(int i=0;i<ct;i++)
printf("%d ",ans[i]);
}
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