[hihoCoder 1190] 连通性·四：点-双连通分量

题意：求点-双连通分量，输出每条边所在双连通分量中边的最小编号。

先前AC了，修改了一个地方，WA。本地对拍，段错误。以为是dfs爆栈，结果是数组越界……那之前是怎么AC的呢？修正了数组的越界，就AC了。

验证了我的理解：pre[v] < pre[u]用来确保反向边只在第一次访问时入栈，换成判断bccno[i->id]是否为0效果是相同的。

#include <cstdio>
#include <vector>
#include <stack>
#include <algorithm>
const int MAX_N = 20000, MAX_M = 100000;
using namespace std;
struct Edge {
    int v, id;
};
typedef vector<Edge>::iterator iter;
vector<Edge> E[MAX_N+1];
stack<Edge> S;
int dfs_clock, bcc_cnt, pre[MAX_N+1], low[MAX_N+1], bccno[MAX_M+1], r[MAX_M+1];

void tarjan(int u, int fa)
{
    low[u] = pre[u] = ++dfs_clock;
    for (iter i = E[u].begin(); i != E[u].end(); ++i) {
        int v = i->v;
        if (!pre[v]) {
            S.push(*i);
            tarjan(v, u);
            if (low[v] >= pre[u]) {
                Edge e;
                ++bcc_cnt;
                do {
                    e = S.top();
                    S.pop();
                    bccno[e.id] = bcc_cnt;
                } while (e.id != i->id);
            } else
                low[u] = min(low[u], low[v]);
        } else if (pre[v] < pre[u] && v != fa) { // 反向边
            S.push(*i);
            low[u] = min(low[u], pre[v]);
        }
    }
}

int main()
{
    int n, m;
    scanf("%d %d", &n, &m);
    for (int i = 1; i <= m; ++i) {
        int u, v;
        scanf("%d %d", &u, &v);
        E[u].push_back((Edge){v, i});
        E[v].push_back((Edge){u, i});
        r[i] = m;
    }
    for (int i = 1; i <= n; ++i)
        if (!pre[i])
            tarjan(i, 0);
    for (int i = 1; i <= m; ++i)
        r[bccno[i]] = min(r[bccno[i]], i);
    printf("%d\n", bcc_cnt);
    for (int i = 1; i <= m; ++i)
        printf("%d%c", r[bccno[i]], " \n"[i==m]);
    return 0;
}