HDU 3836 - Equivalent Sets【强连通分量 基础题】
Description
To prove two sets A and B are equivalent, we can first prove A is a subset of B, and then prove B is a subset of A, so finally we got that these two sets are equivalent.
You are to prove N sets are equivalent, using the method above: in each step you can prove a set X is a subset of another set Y, and there are also some sets that are already proven to be subsets of some other sets.
Now you want to know the minimum steps needed to get the problem proved.
You are to prove N sets are equivalent, using the method above: in each step you can prove a set X is a subset of another set Y, and there are also some sets that are already proven to be subsets of some other sets.
Now you want to know the minimum steps needed to get the problem proved.
Input
The input file contains multiple test cases, in each case, the first line contains two integers N <= 20000 and M <= 50000.
Next M lines, each line contains two integers X, Y, means set X in a subset of set Y.
Next M lines, each line contains two integers X, Y, means set X in a subset of set Y.
Output
For each case, output a single integer: the minimum steps needed.
Sample Input
4 0 3 2 1 2 1 3
Sample Output
4 2
Hint
Case 2: First prove set 2 is a subset of set 1 and then prove set 3 is a subset of set 1.
特别特别水的强连通分量 关于强连通分量的讲解参考这篇博客，都是很裸的== 敲完刘汝佳书上二档模板就1A了
#include<cstdio>
#include<cstring>
#include<vector>
#include<stack>
using namespace std;
#define maxn 20000
vector<int>G[maxn];
int pre[maxn],lowlink[maxn],sccno[maxn],dfs_clock,scc_cnt;
stack<int>S;
int min(int a,int b){if(a<b)return a;return b;}
void dfs(int u)
{
pre[u]=lowlink[u]=++dfs_clock;
S.push(u);
for(int i=0;i<G[u].size();i++)
{
int v=G[u][i];
if(!pre[v])
{
dfs(v);
lowlink[u]=min(lowlink[u],lowlink[v]);
}
else if(!sccno[v])
lowlink[u]=min(lowlink[u],pre[v]);
}
if(lowlink[u]==pre[u])
{
scc_cnt++;
for(;;)
{
int x=S.top();S.pop();
sccno[x]=scc_cnt;
if(x==u) break;
}
}
}
void find_scc(int n)
{
dfs_clock=scc_cnt=0;
memset(sccno,0,sizeof(sccno));
memset(pre,0,sizeof(pre));
for(int i=0;i<n;i++)
if(!pre[i]) dfs(i);
}
int in0[maxn],out0[maxn];
int main()
{
int T,n,m;
while(~scanf("%d%d",&n,&m))
{
for(int i=0;i<n;i++) G[i].clear();
for(int i=0;i<m;i++)
{
int u,v;
scanf("%d%d",&u,&v);
u--;v--;
G[u].push_back(v);
}
find_scc(n);
for(int i=1;i<=scc_cnt;i++)in0[i]=out0[i]=1;
for(int u=0;u<n;u++)
{
for(int i=0;i<G[u].size();i++)
{
int v=G[u][i];
if(sccno[u]!=sccno[v])in0[sccno[v]]=out0[sccno[u]]=0;
}
}
int a=0,b=0;
for(int i=1;i<=scc_cnt;i++)
{
if(in0[i])a++;
if(out0[i]) b++;
}
int ans=a;
if(ans<b) ans=b;
if(scc_cnt==1) ans=0;
printf("%d\n",ans);
}return 0;
}
