UVA-10765 Doves and bombs 【双连通分量】

题目链接：https://vjudge.net/problem/UVA-10765

题目大意：一个无向图中，求去掉每个点后的连通分量的数量。

题解：

这题实际上是求割顶，记录一下割顶的子孙当中反向边不在它之上的连通分量数量，最后加上图初始的连通分量数量。

代码：

#include<bits/stdc++.h>
using namespace std;
#define M(a, b) memset(a, b, sizeof(a))
#define INF 0x3f3f3f3f
const int N = 1e5 + 5;
int pre[N], dfs_clock, tot;
vector<int> G[N];
struct Node {
    int id, val;

    bool operator < (const Node &rhs) const {
        if (rhs.val == val) return id < rhs.id;
        return val > rhs.val;
    }
};
vector<Node> ans;

int dfs(int u, int fa) {
    int lowu = pre[u] = ++dfs_clock;
    int child = 0, num = 0;
    for (int i = 0; i < G[u].size(); ++i) {
        int v = G[u][i];
        if (v == fa) continue;
        if (!pre[v]) {
            ++child;
            int lowv = dfs(v, u);
            lowu = min(lowu, lowv);
            if (lowv >= pre[u]) ++num;
        }
        else lowu = min(lowu, pre[v]);
    }
    if (fa == -1 && child == 1) num = 0;
    ans.push_back(Node{u, num});
    return lowu;
}

void find_bcc(int n) {
    M(pre, 0); dfs_clock = tot = 0;
    ans.clear();
    for (int i = 0; i < n; ++i)
        if (!pre[i]) {
            ++tot;
            dfs(i, -1);
        }
    sort(ans.begin(), ans.end());
}

int main() {
    int n, m;
    while (~scanf("%d%d", &n, &m), n && m) {
        for (int i = 0; i < n; ++i) G[i].clear();
        int u, v;
        while (true) {
            scanf("%d%d", &u, &v);
            if (u == -1 && v == -1) break;
            G[u].push_back(v);
            G[v].push_back(u);
        }
        find_bcc(n);
        for (int i = 0; i < m; ++i) printf("%d %d
", ans[i].id, ans[i].val+tot);
        printf("
");
    }

    return 0;
}