POJ 3352 Road Construction

题目链接：https://vjudge.net/problem/POJ-3352

题目大意

 　　给定一个无向图，问至少添加多少条边，能使无向图变成边双连通图（任意两个节点中，存在两条以上的路径，且路径上的边互不重复）。

分析

　　Tarjan 算法求度为 1 的 e-DCC（边双连通分量） 的模板题。

　　先进行缩点，缩完之后所有的 e-DCC 形成一棵树，然后统计度为 1 的 e-DCC 个数，设为 k，答案就为$frac{k + 1}{2}$。

　　这是因为，如果给两个叶子节点连上线，那么这两个点除了经由根节点相连这条路，又加上直连这条路（产生了回路），如此一来就有 2 条不同的路了，然后既然叶子结点如此，那么叶子结点的父节点也就顺道解决了。因此，只要给叶节点连线即可，最后分奇偶讨论一下。

代码如下

#include <cmath>
#include <ctime>
#include <iostream>
#include <string>
#include <vector>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <queue>
#include <map>
#include <set>
#include <algorithm>
#include <cctype>
#include <stack>
#include <deque>
#include <list>
#include <sstream>
#include <cassert>
using namespace std;
 
#define INIT() ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define Rep(i,n) for (int i = 0; i < (n); ++i)
#define For(i,s,t) for (int i = (s); i <= (t); ++i)
#define rFor(i,t,s) for (int i = (t); i >= (s); --i)
#define ForLL(i, s, t) for (LL i = LL(s); i <= LL(t); ++i)
#define rForLL(i, t, s) for (LL i = LL(t); i >= LL(s); --i)
#define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)
#define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i)
 
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
 
#define LOWBIT(x) ((x)&(-x))
 
#define ALL(x) x.begin(),x.end()
#define INS(x) inserter(x,x.begin())
#define UNIQUE(x) x.erase(unique(x.begin(), x.end()), x.end())
#define REMOVE(x, c) x.erase(remove(x.begin(), x.end(), c), x.end()); // 删去 x 中所有 c 
#define TOLOWER(x) transform(x.begin(), x.end(), x.begin(),::tolower);
#define TOUPPER(x) transform(x.begin(), x.end(), x.begin(),::toupper);
 
#define ms0(a) memset(a,0,sizeof(a))
#define msI(a) memset(a,inf,sizeof(a))
#define msM(a) memset(a,-1,sizeof(a))

#define MP make_pair
#define PB push_back
#define ft first
#define sd second
 
template<typename T1, typename T2>
istream &operator>>(istream &in, pair<T1, T2> &p) {
    in >> p.first >> p.second;
    return in;
}
 
template<typename T>
istream &operator>>(istream &in, vector<T> &v) {
    for (auto &x: v)
        in >> x;
    return in;
}
 
template<typename T1, typename T2>
ostream &operator<<(ostream &out, const std::pair<T1, T2> &p) {
    out << "[" << p.first << ", " << p.second << "]" << "
";
    return out;
}

inline int gc(){
    static const int BUF = 1e7;
    static char buf[BUF], *bg = buf + BUF, *ed = bg;
    
    if(bg == ed) fread(bg = buf, 1, BUF, stdin);
    return *bg++;
} 

inline int ri(){
    int x = 0, f = 1, c = gc();
    for(; c<48||c>57; f = c=='-'?-1:f, c=gc());
    for(; c>47&&c<58; x = x*10 + c - 48, c=gc());
    return x*f;
}

template<class T>
inline string toString(T x) {
    ostringstream sout;
    sout << x;
    return sout.str();
}

inline int toInt(string s) {
    int v;
    istringstream sin(s);
    sin >> v;
    return v;
}

//min <= aim <= max
template<typename T>
inline bool BETWEEN(const T aim, const T min, const T max) {
    return min <= aim && aim <= max;
}
 
typedef long long LL;
typedef unsigned long long uLL;
typedef pair< double, double > PDD;
typedef pair< int, int > PII;
typedef pair< int, PII > PIPII;
typedef pair< string, int > PSI;
typedef pair< int, PSI > PIPSI;
typedef set< int > SI;
typedef set< PII > SPII;
typedef vector< int > VI;
typedef vector< double > VD;
typedef vector< VI > VVI;
typedef vector< SI > VSI;
typedef vector< PII > VPII;
typedef map< int, int > MII;
typedef map< LL, int > MLLI;
typedef map< int, string > MIS;
typedef map< int, PII > MIPII;
typedef map< PII, int > MPIII;
typedef map< string, int > MSI;
typedef map< string, string > MSS;
typedef map< PII, string > MPIIS;
typedef map< PII, PII > MPIIPII;
typedef multimap< int, int > MMII;
typedef multimap< string, int > MMSI;
//typedef unordered_map< int, int > uMII;
typedef pair< LL, LL > PLL;
typedef vector< LL > VL;
typedef vector< VL > VVL;
typedef priority_queue< int > PQIMax;
typedef priority_queue< int, VI, greater< int > > PQIMin;
const double EPS = 1e-8;
const LL inf = 0x3fffffff;
const LL infLL = 0x3fffffffffffffffLL;
const LL mod = 20100713;
const int maxN = 1e3 + 7;
const LL ONE = 1;
const LL evenBits = 0xaaaaaaaaaaaaaaaa;
const LL oddBits = 0x5555555555555555;

struct Edge{
    int from, to;
    
    Edge() {}
    Edge(int x, int y) : from(x), to(y) {}
};

istream& operator>> (istream& in, Edge &x) {
    in >> x.from >> x.to;
    return in;
}

int N, M, ans;
VI V[maxN];
vector< Edge > E;

void addEdge(Edge &x) {
    V[x.from].PB(E.size());
    E.PB(x);
}

stack< int > sk; // 递归处理 SCC (强连通分量)
bool insk[maxN]; // 是否在栈中 

int scc[maxN], sccid; // 存每个点对应SCC的编号 
int in[maxN], out[maxN]; // SCC的入度与出度 

int Time;
int tp[maxN]; // timestamp，时间戳
int facr[maxN]; // The farthest ancestor that can be reached，每个节点最远的返回的祖先 
// S ：当前节点号
// 划分SCC并缩点 
void Tarjan(int S, int fa) {
    tp[S] = facr[S] = ++Time;
    sk.push(S);
    insk[S] = 1;
    
    Rep(i, V[S].size()) {
        Edge &e = E[V[S][i]];
        if(e.to == fa) continue;
        
        if(!tp[e.to]) {
            Tarjan(e.to, S);
            facr[S] = min(facr[S], facr[e.to]);
        }
        else if(insk[e.to]) facr[S] = min(facr[S], tp[e.to]); // 必须要保证在栈中，不然不能保证是一块SCC 
    }
    
    if(facr[S] == tp[S]) {
        ++sccid;
        while(!sk.empty()) {
            int tmp = sk.top(); sk.pop();
            
            insk[tmp] = 0;
            scc[tmp] = sccid;
            if(tmp == S) break;
        } 
    }
}

void init() {
    For(i, 1, N) V[i].clear();
    E.clear();
    ans = 0;
    
    while(!sk.empty()) sk.pop();
    ms0(insk);
    
    ms0(scc); sccid = 0;
    //ms0(in); 
    ms0(out);
    
    Time = 0;
    ms0(tp);
    ms0(facr);
}

int main(){
    //freopen("MyOutput.txt","w",stdout);
    //freopen("input.txt","r",stdin);
    INIT();
    while(cin >> N && N) {
        //init();
        cin >> M;
        For(i, 1, M) {
            Edge t;
            cin >> t;
            addEdge(t);
            swap(t.from, t.to);
            addEdge(t);
        }
        
        For(i, 1, N) if(!tp[i]) Tarjan(i, 0);
        
        For(i, 1, N) {
            Rep(j, V[i].size()) {
                Edge &e = E[V[i][j]];
                if(scc[e.from] != scc[e.to]) {
                    ++out[scc[e.from]];
                    //++in[scc[e.to]];
                }
            }
        }
        
        For(i, 1, sccid) if(out[i] == 1) ++ans; // 因为加了回边，所以只要判断入度或出度 
        
        cout << ((ans + 1) >> 1) << endl;
    }
    return 0;
}