UVA - 10765 Doves and bombs(双连通分量)

题目大意：给定一个n个点的连通的无向图，一个点的“鸽子值“定义为将它从图中删去后连通块的个数，求每个点的“鸽子值“

解题思路：双连通分量的裸题。

#include 
#include 
#include 
using namespace std;
#define N 10010
#define M 100010

struct Edge{
    int to, next;
}E[M];

struct Node{
    int id, times;
}node[N];

int head[N], pre[N];
int n, m, tot, dfs_clock;

bool cmp(const Node &a, const Node &b) {
    if (a.times == b.times)
        return a.id < b.id;
    return a.times > b.times;
}

int dfs(int u, int  fa) {
    int lowu = pre[u] = ++dfs_clock; 
    int child = 0;
    for (int i = head[u]; i != -1; i = E[i].next) {
        int v = E[i].to;
        if (!pre[v]) {
            child++;
            int lowv = dfs(v, u);
            lowu = min(lowu, lowv);
            if (lowv >= pre[u]) {
                node[u].times++;
            }
        }
        else if(pre[v] < pre[u] && v != fa) {
            lowu = min(lowu, pre[v]);
        }
    }

    if (fa < 0 && child == 1)
        node[u].times = 1;
    return lowu;
}

void solve() {
    for (int i = 0; i < n; i++) {
        node[i].id = i;
        node[i].times = 1;
        pre[i] = 0;
    }
    dfs_clock = 0;
    dfs(0, -1);
    sort(node, node + n, cmp);
    for (int i = 0; i < m; i++)
        printf("%d %d\n", node[i].id, node[i].times);
    printf("\n");
}

void AddEdge(int from, int to) {
    E[tot].to = to;
    E[tot].next = head[from];
    head[from] = tot++;
}

void init() {
    memset(head, -1, sizeof(head));
    tot = 0;

    int u, v;
    while (scanf("%d%d", &u, &v) && u >= 0) {
        AddEdge(u, v);
        AddEdge(v, u);
    }
}

int main() {
    while (scanf("%d%d", &n, &m) != EOF && n + m) {
        init();
        solve();
    }
    return 0;
}