UVa 12167 & HDU 2767 强连通分量 Proving Equivalences

题意：给出一个有向图，问最少添加几条有向边使得原图强连通。

解法：求出SCC后缩点，统计一下出度为0的点和入度为0的点，二者取最大值就是答案。

还有个特殊情况就是本身就是强连通的话，答案就是0.

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <stack>
using namespace std;

const int maxn = 20000 + 10;

int n, m;

vector<int> G[maxn];

stack<int> S;
int pre[maxn], low[maxn], sccno[maxn];
int dfs_clock, scc_cnt;

void dfs(int u, int fa)
{
    low[u] = pre[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, u);
            low[u] = min(low[u], low[v]);
        }
        else if(!sccno[v]) low[u] = min(low[u], pre[v]);
    }

    if(pre[u] == low[u])
    {
        scc_cnt++;
        for(;;)
        {
            int x = S.top(); S.pop();
            sccno[x] = scc_cnt;
            if(x == u) break;
        }
    }
}

void find_scc()
{
    memset(pre, 0, sizeof(pre));
    memset(sccno, 0, sizeof(sccno));
    dfs_clock = scc_cnt = 0;
    for(int i = 1; i <= n; i++) if(!pre[i]) dfs(i, 0);
}

int in[maxn], out[maxn];

int main()
{
    int T; scanf("%d", &T);
    while(T--)
    {
        scanf("%d%d", &n, &m);
        for(int i = 1; i <= n; i++) G[i].clear();
        while(m--)
        {
            int u, v; scanf("%d%d", &u, &v);
            G[u].push_back(v);
        }

        find_scc();

        if(scc_cnt == 1) { puts("0"); continue; }

        memset(in, 0, sizeof(in));
        memset(out, 0, sizeof(out));
        for(int i = 1; i <= n; i++)
            for(int j = 0; j < G[i].size(); j++)
            {
                int u = sccno[i], v = sccno[G[i][j]];
                if(u != v) { out[u]++; in[v]++; }
            }

        int hehe = 0, haha = 0;
        for(int i = 1; i <= scc_cnt; i++)
        {
            if(!in[i]) hehe++;
            if(!out[i]) haha++;
        }
        printf("%d
", max(hehe, haha));
    }

    return 0;
}