UVa 1719 Tours 2015 WF

方法：桥

2015 WF的题，很巧妙，看过题解恍然大悟。这道题本质是将一个无向图的边用k种颜色染色，使得对于每一个环，这个环上每种颜色的边，数量相同。首先，我们说一个intuition，如果我们把每个环的长度都求出来并且求gcd，那么这个gcd应该是最大长度的倍数。

code:

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <string>
#include <vector>
#include <stack>
#include <bitset>
#include <cstdlib>
#include <cmath>
#include <set>
#include <list>
#include <deque>
#include <map>
#include <queue>
#include <fstream>
#include <cassert>
#include <unordered_map>
#include <unordered_set>
#include <cmath>
#include <sstream>
#include <time.h>
#include <complex>
#include <iomanip>
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
#define FOR(a,b,c) for (ll (a)=(b);(a)<(c);++(a))
#define FORN(a,b,c) for (ll (a)=(b);(a)<=(c);++(a))
#define DFOR(a,b,c) for (ll (a)=(b);(a)>=(c);--(a))
#define FORSQ(a,b,c) for (ll (a)=(b);(a)*(a)<=(c);++(a))
#define FORC(a,b,c) for (char (a)=(b);(a)<=(c);++(a))
#define FOREACH(a,b) for (auto &(a) : (b))
#define rep(i,n) FOR(i,0,n)
#define repn(i,n) FORN(i,1,n)
#define drep(i,n) DFOR(i,n-1,0)
#define drepn(i,n) DFOR(i,n,1)
#define MAX(a,b) a = Max(a,b)
#define MIN(a,b) a = Min(a,b)
#define SQR(x) ((LL)(x) * (x))
#define Reset(a,b) memset(a,b,sizeof(a))
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define all(v) v.begin(),v.end()
#define ALLA(arr,sz) arr,arr+sz
#define SIZE(v) (int)v.size()
#define SORT(v) sort(all(v))
#define REVERSE(v) reverse(ALL(v))
#define SORTA(arr,sz) sort(ALLA(arr,sz))
#define REVERSEA(arr,sz) reverse(ALLA(arr,sz))
#define PERMUTE next_permutation
#define TC(t) while(t--)
#define forever for(;;)
#define PINF 1000000000000
#define newline '
'

#define test if(1)if(0)cerr
using namespace std;
using namespace std;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int,int> ii;
typedef pair<double,double> dd;
typedef pair<char,char> cc;
typedef vector<ii> vii;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll> l4;
const double pi = acos(-1.0);


int n, m, ans;
const int maxn = 2000+1;
set<int> g[maxn];

int dfn[maxn], low[maxn], dfs_clcok;
bool vis[maxn][maxn];
vector<ii> ret;
void bridges(int cur, int pa, int &times)
{
    dfn[cur] = low[cur] = times;
    for (const int &nxt : g[cur])
    {
        if (nxt == pa) continue;
        if (!dfn[nxt])
        {
            bridges(nxt, cur, ++times);
            low[cur] = min(low[cur], low[nxt]);
            if (low[nxt] == dfn[nxt]) ret.pb(mp(cur, nxt));
        }
        else
            low[cur] = min(low[cur], dfn[nxt]);
    }
}
void solve()
{
    ret.clear();
    Reset(dfn, 0);
    repn(i, n) if (!dfn[i]) bridges(i, -1, dfs_clcok=1);
    for (int i = 0; i < ret.size(); ++i)
        vis[ret[i].first][ret[i].second] = vis[ret[i].second][ret[i].first] = true;

}
int gcd(int a, int b)
{
    return b?gcd(b, a%b):a;
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    while (cin >> n >> m)
    {
        Reset(vis, false);
        repn(i, n) g[i].clear();
        rep(i, m)
        {
            int u, v;   cin >> u >> v;
            g[u].insert(v);
            g[v].insert(u);
        }
        ans = -1;
        solve();
        for (int i = 0; i < ret.size(); ++i)
        {
            int u = ret[i].first, v = ret[i].second;
            g[u].erase(v);
            g[v].erase(u);
        }
        repn(i, n)
        {
            auto temp = g[i];
            for (auto nxt : temp)
            {
                if (vis[i][nxt]) continue;
                vis[i][nxt] = vis[nxt][i] = true;
                g[i].erase(nxt);
                g[nxt].erase(i);
                solve();
                if (ans == -1) ans = ret.size()+1;
                else ans = gcd(ans, ret.size()+1);
                g[i].insert(nxt);
                g[nxt].insert(i);
            }
        }
        if (ans == -1)
        {
            
            repn(i, m-1) cout << i << " ";
            cout << m;
            cout << newline;
        }
        else
        {
            repn(i, ans-1) if (ans % i == 0) cout << i << " ";
            cout << ans;
            cout << newline;
        }
    }
}
/*
 4 5
 1 2 
 2 3 
 3 4 
 1 4 
 1 3 
 
 6 6 
 1 2 
 2 3 
 1 3 
 1 4 
 2 5 
 3 6
*/