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:
1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 #include <iostream> 5 #include <string> 6 #include <vector> 7 #include <stack> 8 #include <bitset> 9 #include <cstdlib> 10 #include <cmath> 11 #include <set> 12 #include <list> 13 #include <deque> 14 #include <map> 15 #include <queue> 16 #include <fstream> 17 #include <cassert> 18 #include <unordered_map> 19 #include <unordered_set> 20 #include <cmath> 21 #include <sstream> 22 #include <time.h> 23 #include <complex> 24 #include <iomanip> 25 #define Max(a,b) ((a)>(b)?(a):(b)) 26 #define Min(a,b) ((a)<(b)?(a):(b)) 27 #define FOR(a,b,c) for (ll (a)=(b);(a)<(c);++(a)) 28 #define FORN(a,b,c) for (ll (a)=(b);(a)<=(c);++(a)) 29 #define DFOR(a,b,c) for (ll (a)=(b);(a)>=(c);--(a)) 30 #define FORSQ(a,b,c) for (ll (a)=(b);(a)*(a)<=(c);++(a)) 31 #define FORC(a,b,c) for (char (a)=(b);(a)<=(c);++(a)) 32 #define FOREACH(a,b) for (auto &(a) : (b)) 33 #define rep(i,n) FOR(i,0,n) 34 #define repn(i,n) FORN(i,1,n) 35 #define drep(i,n) DFOR(i,n-1,0) 36 #define drepn(i,n) DFOR(i,n,1) 37 #define MAX(a,b) a = Max(a,b) 38 #define MIN(a,b) a = Min(a,b) 39 #define SQR(x) ((LL)(x) * (x)) 40 #define Reset(a,b) memset(a,b,sizeof(a)) 41 #define fi first 42 #define se second 43 #define mp make_pair 44 #define pb push_back 45 #define all(v) v.begin(),v.end() 46 #define ALLA(arr,sz) arr,arr+sz 47 #define SIZE(v) (int)v.size() 48 #define SORT(v) sort(all(v)) 49 #define REVERSE(v) reverse(ALL(v)) 50 #define SORTA(arr,sz) sort(ALLA(arr,sz)) 51 #define REVERSEA(arr,sz) reverse(ALLA(arr,sz)) 52 #define PERMUTE next_permutation 53 #define TC(t) while(t--) 54 #define forever for(;;) 55 #define PINF 1000000000000 56 #define newline ‘\n‘ 57 58 #define test if(1)if(0)cerr 59 using namespace std; 60 using namespace std; 61 typedef vector<int> vi; 62 typedef vector<vi> vvi; 63 typedef pair<int,int> ii; 64 typedef pair<double,double> dd; 65 typedef pair<char,char> cc; 66 typedef vector<ii> vii; 67 typedef long long ll; 68 typedef unsigned long long ull; 69 typedef pair<ll, ll> l4; 70 const double pi = acos(-1.0); 71 72 73 int n, m, ans; 74 const int maxn = 2000+1; 75 set<int> g[maxn]; 76 77 int dfn[maxn], low[maxn], dfs_clcok; 78 bool vis[maxn][maxn]; 79 vector<ii> ret; 80 void bridges(int cur, int pa, int ×) 81 { 82 dfn[cur] = low[cur] = times; 83 for (const int &nxt : g[cur]) 84 { 85 if (nxt == pa) continue; 86 if (!dfn[nxt]) 87 { 88 bridges(nxt, cur, ++times); 89 low[cur] = min(low[cur], low[nxt]); 90 if (low[nxt] == dfn[nxt]) ret.pb(mp(cur, nxt)); 91 } 92 else 93 low[cur] = min(low[cur], dfn[nxt]); 94 } 95 } 96 void solve() 97 { 98 ret.clear(); 99 Reset(dfn, 0); 100 repn(i, n) if (!dfn[i]) bridges(i, -1, dfs_clcok=1); 101 for (int i = 0; i < ret.size(); ++i) 102 vis[ret[i].first][ret[i].second] = vis[ret[i].second][ret[i].first] = true; 103 104 } 105 int gcd(int a, int b) 106 { 107 return b?gcd(b, a%b):a; 108 } 109 int main() 110 { 111 ios::sync_with_stdio(false); 112 cin.tie(0); 113 while (cin >> n >> m) 114 { 115 Reset(vis, false); 116 repn(i, n) g[i].clear(); 117 rep(i, m) 118 { 119 int u, v; cin >> u >> v; 120 g[u].insert(v); 121 g[v].insert(u); 122 } 123 ans = -1; 124 solve(); 125 for (int i = 0; i < ret.size(); ++i) 126 { 127 int u = ret[i].first, v = ret[i].second; 128 g[u].erase(v); 129 g[v].erase(u); 130 } 131 repn(i, n) 132 { 133 auto temp = g[i]; 134 for (auto nxt : temp) 135 { 136 if (vis[i][nxt]) continue; 137 vis[i][nxt] = vis[nxt][i] = true; 138 g[i].erase(nxt); 139 g[nxt].erase(i); 140 solve(); 141 if (ans == -1) ans = ret.size()+1; 142 else ans = gcd(ans, ret.size()+1); 143 g[i].insert(nxt); 144 g[nxt].insert(i); 145 } 146 } 147 if (ans == -1) 148 { 149 150 repn(i, m-1) cout << i << " "; 151 cout << m; 152 cout << newline; 153 } 154 else 155 { 156 repn(i, ans-1) if (ans % i == 0) cout << i << " "; 157 cout << ans; 158 cout << newline; 159 } 160 } 161 } 162 /* 163 4 5 164 1 2 165 2 3 166 3 4 167 1 4 168 1 3 169 170 6 6 171 1 2 172 2 3 173 1 3 174 1 4 175 2 5 176 3 6 177 */
原文:http://www.cnblogs.com/skyette/p/6359152.html