hdoj 3836 Equivalent Sets

题目链接：Equivalent Sets

题目大意：要证明两个集合A、B等价，需要得到X是Y的子集并且Y也是X的子集，现在有N个集合，M个条件，代表a是b的子集，问最少需要再添加多少条件使得所有N个集合都是等价的

题目思路：X是Y的子集，也就是Y到X有一条有向边，那么题目就转换为了给你一个有向图，问最少添加多少条边使得所有的点都能到达任何另外的点，tarjan缩点，找出度为0和入度为0的最小值就可以了

#include <map>#include <set>#include <cmath>#include <stack>#include <queue>#include <vector>#include <cstdio>#include <string>#include <utility>#include <cstdlib>#include <cstring>#include <iostream>#include <algorithm>using namespace std;typedef long long ll;const int maxn = 20010;vector<int>Edge[maxn];stack<int>S;int Dfn[maxn],Low[maxn],sccno[maxn],tclock,sccnt;int InDeg[maxn],OutDeg[maxn];void tarjan(int u){    Dfn[u] = Low[u] = ++tclock;    S.push(u);    for(int i = 0;i < Edge[u].size();i++){        int v = Edge[u][i];        if(!Dfn[v]){            tarjan(v);            Low[u] = min(Low[u],Low[v]);        }        else if(!sccno[v]){            Low[u] = min(Low[u],Dfn[v]);        }    }    if(Low[u] == Dfn[u]){        sccnt += 1;        int v = -1;        while(v != u){            v = S.top();            S.pop();            sccno[v] = sccnt;        }    }}void findscc(int n){    tclock = sccnt = 0;    memset(Dfn,0,sizeof(Dfn));    memset(Low,0,sizeof(Low));    memset(sccno,0,sizeof(sccno));    for(int i = 1;i <= n;i++)        if(!Dfn[i]) tarjan(i);}int solve(int n){    if(sccnt == 1) return 0;    memset(InDeg,0,sizeof(InDeg));    memset(OutDeg,0,sizeof(OutDeg));    for(int u = 1;u <= n;u++){        for(int i = 0;i < Edge[u].size();i++){            int v = Edge[u][i];            if(sccno[u] != sccno[v]){                InDeg[sccno[v]]++;                OutDeg[sccno[u]]++;            }        }    }    int c1 = 0,c2 = 0;    for(int i = 1;i <= sccnt;i++){        if(InDeg[i] == 0) c1++;        if(OutDeg[i] == 0) c2++;    }    return max(c1,c2);}int main(){    int T,n,m;    while(~scanf("%d%d",&n,&m)){        for(int i = 1;i <= n;i++) Edge[i].clear();        while(m--){            int u,v;            scanf("%d%d",&u,&v);            Edge[v].push_back(u);        }        findscc(n);        printf("%d\n",solve(n));    }}