2015
04-13

# Equivalent Sets

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.

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.

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.

4 0
3 2
1 2
1 3

4
2
HintCase 2: First prove set 2 is a subset of set 1 and then prove set 3 is a subset of set 1. 

/*

Tarjan+缩点。
Tarjan求出有几个强连通分量，如果是1的话，那么ans=0；

2012-10-29
*/

#include"stdio.h"
#include"string.h"
#include"stdlib.h"
#include"stack"
#define N 30000
using namespace std;
int n,m;
int index_s;
int instack[N],DFN[N],LOW[N];
int belong[N],indegree[N],outdegree[N];
struct Eage
{
int from,to,next;
}eage[2*N];

void add(int a,int b)
{
eage[tot].from=a;
eage[tot].to=b;
}
void getmap()
{
int i,l;
int a,b;
tot=0;
}
stack<int>st;
void Tarjan(int k)
{
int j,v;
st.push(k);
instack[k]=1;
DFN[k]=LOW[k]=++index_s;
{
v=eage[j].to;
if(instack[v])  LOW[k]=LOW[k]>DFN[v]?DFN[v]:LOW[k];
else if(DFN[v]==-1)
{
Tarjan(v);
LOW[k]=LOW[k]>LOW[v]?LOW[v]:LOW[k];
}
}
if(DFN[k]==LOW[k])
{
do
{
j=st.top();
st.pop();
instack[j]=0;
belong[j]=k;
}while(j!=k);
}
}
void getdegree()
{
int i,l;
memset(indegree,0,sizeof(indegree));
memset(outdegree,0,sizeof(outdegree));
for(i=0;i<tot;i++)
{
if(belong[eage[i].from]==belong[eage[i].to])    continue;
indegree[belong[eage[i].to]]++;
outdegree[belong[eage[i].from]]++;
}
}
int main()
{
int i;
int temp,t1,t2,ans;
while(scanf("%d%d",&n,&m)!=-1)
{
getmap();

index_s=0;
memset(DFN,-1,sizeof(DFN));
memset(LOW,-1,sizeof(LOW));
memset(instack,0,sizeof(instack));
for(i=1;i<=n;i++)   if(DFN[i]==-1) Tarjan(i);

getdegree();

temp=t1=t2=0;
for(i=1;i<=n;i++)
{
if(belong[i]!=i)    continue;
temp++;
if(indegree[i]==0)  t1++;
if(outdegree[i]==0) t2++;
}
ans=t1>t2?t1:t2;
if(n<1 || temp==1) ans=0;
printf("%d\n",ans);
}
return 0;
}

,
1. 5.1处，反了；“上一个操作符的优先级比操作符ch的优先级大，或栈是空的就入栈。”如代码所述，应为“上一个操作符的优先级比操作符ch的优先级小，或栈是空的就入栈。”