点的双连通分量问题

连通性·四

题目传送：hihoCoder - 1190 - 连通性·四

好吧，这简单题敲了好久。。快一上午了，，基础不好诶。。

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 0x7fffffff
using namespace std;

const int maxn = 20005;
int n, m;
pair<int, int> e[100005];//用pair存边 

vector<int> G[maxn];//整个图 
map<pair<int, int>, int> mp;//标记每条边的分组，刚开始为0 
map<pair<int, int>, int> id;//记录每条边的编号 

int vis[maxn];

int dfn[maxn];
int low[maxn];

stack<pair<int, int> > s;

int cur_time, cnt;
void dfs(int u, int fa) {
    dfn[u] = low[u] = ++ cur_time;
    vis[u] = 1;

    int child = 0;
    int d = G[u].size();
    for(int i = 0; i < d; i ++) {

        int v = G[u][i];
        pair<int, int> pai = make_pair(min(u, v), max(u, v));

        if(mp[pai] != 0) continue;
        if(!vis[v]) {
            child ++;
            s.push(pai);
            dfs(v, u);
            low[u] = min(low[u], low[v]);

            if(fa == -1 && child > 1) {//求出树根为割点的时候的点的双连通分量 
                cnt ++;
                stack<pair<int, int> > ss;
                int MIN = INF;
                while(s.top() != pai) {
                    MIN = min(MIN, id[s.top()]);
                    ss.push(s.top());
                    s.pop();
                }
                MIN = min(MIN, id[pai]);
                ss.push(pai);
                s.pop();

                //cout << MIN << " haha";
                while(!ss.empty()) {
                    mp[ss.top()] = MIN;
                    ss.pop();
                }
            }

            if(fa != -1 && low[v] >= dfn[u]) {//求出树根以外的结点为割点的点的双连通分量 
                //cout << "low " << low[v] << " dfn " << dfn[u] << endl; 
                cnt ++;
                stack<pair<int, int> > ss;
                int MIN = INF;
                while(s.top() != pai) {
                    MIN = min(MIN, id[s.top()]);
                    //cout << id[s.top()] << " " << s.top().first << " " << s.top().second << endl;
                    ss.push(s.top());
                    s.pop();
                }
                //cout << id[pai] << endl;
                MIN = min(MIN, id[pai]);
                ss.push(pai);
                s.pop();

                //cout << MIN << " haha" << endl;
                while(!ss.empty()) {
                    mp[ss.top()] = MIN;
                    ss.pop();
                }
            }
        }
        else if(v != fa) {//反向边的情况 
            s.push(pai);
            low[u] = min(low[u], dfn[v]);
        }
    }
}

int main() {
    scanf("%d %d", &n, &m);
    for(int i = 1; i <= m; i ++) {
        scanf("%d %d", &e[i].first, &e[i].second);
        G[e[i].first].push_back(e[i].second);
        G[e[i].second].push_back(e[i].first);
        mp[e[i]] = 0;
        id[e[i]] = i;
    }

    cur_time = cnt = 0;
    dfs(1, -1);

    //求剩余那部分的点的双连通分量 
    stack<pair<int, int> > ss;
    int MIN = INF;
    while(!s.empty()) {
        ss.push(s.top());
        MIN = min(MIN, id[s.top()]);
        s.pop();
    }
    if(MIN != INF) cnt ++;
    while(!ss.empty()) {
        mp[ss.top()] = MIN;
        ss.pop();
    }

    printf("%d\n", cnt);
    for(int i = 1; i < m; i ++) {
        printf("%d ", mp[e[i]]);
    }
    printf("%d\n", mp[e[m]]);
    return 0;
}