边的双连通分量问题

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连通性二·边的双连通分量

题目传送：hihoCoder - 1184 - 连通性二·边的双连通分量

AC代码:

#include <map>#include <set>#include <list>#include <cmath>#include <deque>#include <queue>#include <stack>#include <bitset>#include <cctype>#include <cstdio>#include <string>#include <vector>#include <complex>#include <cstdlib>#include <cstring>#include <fstream>#include <sstream>#include <utility>#include <iostream>#include <algorithm>#include <functional>#define LL long long#define INF 0x7fffffffusing namespace std;const int maxn = 20005;int n, m;vector<int> G[maxn];stack<int> s;int vis[maxn];int low[maxn];int dfn[maxn];int ans[maxn];int cur_time, cnt;void dfs(int u, int fa) {    low[u] = dfn[u] = ++ cur_time;    vis[u] = 1;    s.push(u);    int d = G[u].size();    for(int i = 0; i < d; i ++) {        int v = G[u][i];        if(!vis[v]) {            dfs(v, u);            low[u] = min(low[u], low[v]);        }        else if(v != fa){            low[u] = min(low[u], dfn[v]);           }    }    if(low[u] == dfn[u]) {        cnt ++;        stack<int> ss;        int MIN = INF;        while(s.top() != u) {            MIN = min(MIN, s.top());            ss.push(s.top());            s.pop();        }        MIN = min(MIN, s.top());        ss.push(s.top());        s.pop();        while(!ss.empty()) {            ans[ss.top()] = MIN;            ss.pop();        }    }}int main() {    scanf("%d %d", &n, &m);    int u, v;    for(int i = 0; i < m; i ++) {        scanf("%d %d", &u, &v);        G[u].push_back(v);        G[v].push_back(u);    }    while(!s.empty()) s.pop();    cur_time = cnt = 0;    memset(vis, 0, sizeof(vis));    dfs(1, -1);    printf("%d\n", cnt);    for(int i = 1; i < n; i ++) {        printf("%d ", ans[i]);    }    printf("%d\n", ans[n]);    return 0;}

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