POJ 3177

题目大意：

给定一个N个点，M条边的无向连通图（可能有重边），要求让任意两点间都有两条或以上的路径，且这些路径没有公共边。问至少需要加多少条边？

N<=5e3，M<=1e4。

求双连通分量并缩点。详见：https://www.cnblogs.com/frog112111/p/3367039.html

注意由于本题数据允许重边（尽管讨论区有人吐槽数据太水），DFS时判断割边的条件应为low[to] > dfn[cur]，且该边的重数为1。

实现的时候用了并查集来维护属于同一双联通分量的点，标记割边时将它的重数变成-1。

代码：（都8102年了贵OJ还不支持C++11-_-||）

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <map>

using namespace std;

const int maxN = 5000 + 5;
const int notVisited = -1;

map<int, int> elist[maxN];
int dfn[maxN];
int low[maxN];
int ufs[maxN];
int cnt[maxN];
int lastDfn = 0;
int N, M;

inline void addEdge(int u, int v)
{
    map<int, int>::iterator it = elist[u].find(v);
    if (it != elist[u].end())
        it->second += 1;
    else
        elist[u].insert(make_pair(v, 1));
}

void input()
{
    scanf("%d%d", &N, &M);
    for (int u, v, i = 1; i <= M; i++)
    {
        scanf("%d%d", &u, &v);
        addEdge(u, v);
        addEdge(v, u);
    }
}

void init()
{
    memset(dfn, -1, sizeof(dfn));
    memset(low, 0x3f, sizeof(low)); //infinity
    lastDfn = 0;
    for (int i = 1; i <= N; i++)
        ufs[i] = i;
}

int findUfs(int x)
{
    return ufs[x] == x ? x : (ufs[x] = findUfs(ufs[x]));
}

bool unifyUfs(int x, int y)
{
    int fx = findUfs(x);
    int fy = findUfs(y);
    if (fx == fy)
        return false;
    ufs[fx] = fy;
    return true;
}

void dfs(int cur, int last)
{
    dfn[cur] = low[cur] = (++lastDfn);

    for (map<int, int>::iterator it = elist[cur].begin(); it != elist[cur].end(); ++it)
    {
        int to = it->first;
        if (dfn[to] == notVisited)
        {
            dfs(to, cur);
            if (it->second >= 2 || low[to] <= dfn[cur])
                unifyUfs(cur, to);
            else
                it->second = -1;
            low[cur] = min(low[cur], low[to]);
        }
        else if (to != last)
        {
            unifyUfs(to, cur);
            low[cur] = min(low[cur], dfn[to]);
        }
    }
}

int solve()
{
    init();
    dfs(1, 0);

    for (int i = 1; i <= N; i++)
        for (map<int, int>::iterator it = elist[i].begin(); it != elist[i].end(); ++it)
        {
            if (it->second > 0)
                continue;

            int to = it->first;
            int fi = findUfs(i);
            int ft = findUfs(to);
            if (fi != ft)
            {
                cnt[fi] += 1;
                cnt[ft] += 1;
            }
        }

    return ((int)count(cnt + 1, cnt + N + 1, 1) + 1) / 2;
}

int main()
{
    input();
    printf("%d", solve());
    return 0;
}