hdu 2767 Proving Equivalences（tarjan缩点）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2767

题意：问最少加多少边可以让所有点都相互连通。

题解：如果强连通分量就1个直接输出0，否者输出入度为0的缩点，出度为0的缩点和的最大值

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
const int N = 2e4 + 10;
const int M = 5e4 + 10;
struct Edge {
    int v, next;
}edge[M];
int head[N], e;
int Low[N], DFN[N], Belong[N], num[N], Stack[N];
int Index, top, scc;
int In[N], Out[N];
bool Instack[N];
void init() {
    memset(head, -1, sizeof(head));
    e = 0;
}
void add(int u, int v) {
    edge[e].v = v;
    edge[e].next = head[u];
    head[u] = e++;
}
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].v;
        if(!DFN[v]) {
            Tarjan(v);
            Low[u] = min(Low[u], Low[v]);
        } else if(Instack[v]) Low[u] = min(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);
    }
}
int main() {
    int t;
    scanf("%d", &t);
    while(t--) {
        int n, m;
        scanf("%d%d", &n, &m);
        init();
        for(int i = 0; i < m; i++) {
            int s1, s2;
            scanf("%d%d", &s1, &s2);
            add(s1, s2);
        }
        memset(DFN, 0, sizeof(DFN));
        memset(Instack, false, sizeof(Instack));
        memset(Belong, 0, sizeof(Belong));
        memset(num, 0, sizeof(num));
        memset(In, 0, sizeof(In));
        memset(Out, 0, sizeof(Out));
        scc = 0, Index = 0, top = 0;
        for(int i = 1; i <= n; i++) {
            if(!DFN[i]) Tarjan(i);
        }
        for(int i = 1; i <= n; i++) {
            for(int j = head[i]; j != -1; j = edge[j].next) {
                int v = edge[j].v;
                if(Belong[i] != Belong[v]) {In[Belong[v]]++, Out[Belong[i]]++;}
            }
        }
        int mxin = 0, mxout = 0;
        for(int i = 1; i <= scc; i++) {
            if(In[i] == 0) mxin++;
            if(Out[i] == 0) mxout++;
        }
        if(scc == 1) printf("0\n");
        else printf("%d\n", max(mxin, mxout));
    }
    return 0;
}