题目大意：给出N个点，M条边。要求你加入最少的边，使得这个图变成强连通分量

解题思路：先找出全部的强连通分量和桥，将强连通分量缩点。桥作为连线，就形成了DAG了

这题被坑了。用了G＋＋交的，结果一直RE，用C＋＋一发就过了。。。

#include <cstdio>

#include <cstring>

#define N 20010

#define M 100010

#define min(a,b) ((a) > (b)?

(b): (a))

#define max(a,b) ((a) > (b)?

(a): (b))

struct Edge{

    int from, to, next;

}E[M];

int head[N], sccno[N], pre[N], lowlink[N], stack[N], in[N], out[N];

int n, m, tot, dfs_clock, top, scc_cnt;

void AddEdge(int from, int to) {

    E[tot].from = from;

    E[tot].to = to;

    E[tot].next = head[from];

    head[from] = tot++;

}

void init() {

    memset(head, -1, sizeof(head));

    tot = 0;

    int u, v;

    for (int i = 0; i < m; i++) {

        scanf("%d%d", &u, &v);

        AddEdge(u, v);

    }

}

void dfs(int u) {

    pre[u] = lowlink[u] = ++dfs_clock;

    stack[++top] = u;

    for (int i = head[u]; i != -1; i = E[i].next) {

        int v = E[i].to;

        if (!pre[v]) {

            dfs(v);

            lowlink[u] = min(lowlink[u], lowlink[v]);

        }

        else if (!sccno[v]) {

            lowlink[u] = min(lowlink[u], pre[v]);

        }

    }

    int x;

    if (pre[u] == lowlink[u]) {

        scc_cnt++;

        while (1) {

            x = stack[top--];

            sccno[x] = scc_cnt;

            if (x == u)

                break;

        }

    }

}

void solve() {

    memset(sccno, 0, sizeof(sccno));

    memset(pre, 0, sizeof(pre));

    dfs_clock = top = scc_cnt = 0;

    for (int i = 1; i <= n; i++)

        if (!pre[i])

            dfs(i);

    if (scc_cnt <= 1) {

        printf("0\n");

        return ;

    }

    for (int i = 1; i <= scc_cnt; i++)

        in[i] = out[i] = 1;

    for (int i = 0; i < tot; i++) {

        int u = E[i].from, v = E[i].to;

        if (sccno[u] != sccno[v]) {

            out[sccno[u]] = in[sccno[v]] = 0;

        }

    }

    int a = 0, b = 0;

    for (int i = 1; i <= scc_cnt; i++) {

        if (out[i]) a++;

        if (in[i]) b++;

    }

    printf("%d\n", max(a, b));

}

int main() {

    while (scanf("%d%d", &n, &m) != EOF) {

        init();

        solve();

    }

    return 0;

}