Equivalent Sets

Time Limit: 12000/4000 MS (Java/Others)    Memory Limit: 104857/104857 K (Java/Others) Total Submission(s): 6145    Accepted Submission(s): 2202 Problem 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.

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.

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.

Source

問最少加幾條邊，使整個有向圖強連通

先縮點，縮點之後形成的新圖，出度為0的個數a和入度為0的個數b中較大的值為答案

注意特判該圖本身就為強連通的情況

#include <bits/stdc++.h>
using namespace std;
const int MAXN = 20005;
const int MAXM = 50005;
/*************************************/
struct Edge
{
    int to,Next;
}edge[MAXM];
int head[MAXN],tot;
int Low[MAXN],DFN[MAXN],Stack[MAXN],Belong[MAXN];
int Index,top;
int scc;
bool Instack[MAXN];
int num[MAXN];
void addedge(int u,int v)
{
    edge[tot].to = v;
    edge[tot].Next = head[u];
    head[u] = tot++;
}
void Tarjan(int u)
{
    int v;
    Low[u] = DFN[u] = ++Index;
    Stack[top++] = u;
    Instack[u] = true;
    for(int i = head[u]; i != -1; i = edge[i].Next) {
        v = edge[i].to;
        if(!DFN[v]) {
            Tarjan(v);
            if(Low[u] > Low[v]) Low[u] = Low[v];
        }
        else if(Instack[v] && Low[u] > DFN[v])
            Low[u] = DFN[v];
    }
    if(Low[u] == DFN[u]) {
        scc++;
        do {
            v = Stack[--top];
            Instack[v] = false;
            Belong[v] = scc;
            num[scc]++;
        }
        while(v != u);
    }
}
void solve(int N)
{
    memset(DFN,0,sizeof(DFN));
    memset(Instack,false,sizeof(Instack));
    memset(num,0,sizeof(num));
    Index = scc = top = 0;
    for(int i = 1; i <= N; i++) {
        if(!DFN[i]) {
            Tarjan(i);
        }
    }
}
int in[MAXN],out[MAXN];
void init()
{
    tot = 0;
    memset(in,0,sizeof(in));
    memset(out,0,sizeof(out));
    memset(head,-1,sizeof(head));
}
/************************************/
int main(void)
{
    int n,m;
    int u,v;
    int T;
    scanf("%d",&T);
    while(T--) {
        scanf("%d %d",&n,&m);
        init();
        for(int i = 1; i <= m; i++) {
            scanf("%d %d",&u,&v);
            addedge(u,v);
        }
        solve(n);
        for(int i = 1; i <= n; i++) {
            for(int j = head[i]; j != -1; j = edge[j].Next) {
                u = Belong[i],v = Belong[edge[j].to];
                if(u != v) {
                    out[u]++;
                    in[v]++;
                }
            }
        }
        int ans1 = 0,ans2 = 0;
        for(int i = 1; i <= scc; i++) {
            if(in[i] == 0) ans1++;
            if(out[i] == 0) ans2++;
        }
        printf("%d\n",scc == 1 ? 0 : max(ans1,ans2));
    }
    return 0;
}