HDU - 3836 Equivalent Sets (强连通分量+DAG)

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;
}