基环树专题
最近做了下题,作业题目有一道很有意思的题目
[CF 711D](https://codeforces.com/problemset/problem/711/D)
这道题问的是给出一个必存在至少一个环的图里面,每次操作可以选出来一些边,每条有向边可以被反转,然后问最后能让这个图无环的方案有多少种
其实熟悉基环树的都一眼能看出来这就是基环树,因为n == m代表图中存在一个大环,我们随意反转树上的边是不影响的,所以我们可以选出来0- n - 1条边(树的大小)
这个很显然,有每个子树对答案的贡献是pow(2, 大小 + 1) - 2
然后处理完剩下的自然就是环了,所以把 |图大小 - 子树大小| 即为环,基环上任意一点反转,则环消失变为dag,但不能同时转向(又成环了),所以总数有pow(2, n - tot)种
乘起来取模就好了,妙哇!
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include <bits/stdc++.h> using namespace std; constexpr int limit = (4000000 + 5);//防止溢出 #define INF 0x3f3f3f3f #define inf 0x3f3f3f3f3f #define lowbit(i) i&(-i)//一步两步 #define EPS 1e-9 #define FASTIO ios::sync_with_stdio(false);cin.tie(0),cout.tie(0); #define ff(a) printf("%d\n",a ); #define pi(a, b) pair<a,b> #define rep(i, a, b) for(ll i = a; i <= b ; ++i) #define per(i, a, b) for(ll i = b ; i >= a ; --i) #define MOD 998244353 #define traverse(u) for(int i = head[u]; ~i ; i = edge[i].next) #define FOPEN freopen("C:\\Users\\tiany\\CLionProjects\\akioi\\data.txt", "rt", stdin) #define FOUT freopen("C:\\Users\\tiany\\CLionProjects\\akioi\\dabiao.txt", "wt", stdout) typedef long long ll; typedef unsigned long long ull; char buf[1 << 23], *p1 = buf, *p2 = buf, obuf[1 << 23], *O = obuf; inline ll read() { #define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++) ll sign = 1, x = 0; char s = getchar(); while (s > '9' || s < '0') { if (s == '-')sign = -1; s = getchar(); } while (s >= '0' && s <= '9') { x = (x << 3) + (x << 1) + s - '0'; s = getchar(); } return x * sign; #undef getchar }//快读 void print(ll x) { if (x / 10) print(x / 10); *O++ = x % 10 + '0'; } void write(ll x, char c = 't') { if (x < 0)putchar('-'), x = -x; print(x); if (!isalpha(c))*O++ = c; fwrite(obuf, O - obuf, 1, stdout); O = obuf; } constexpr ll mod = 1e9 + 7; ll quickPow(ll base, ll expo){ ll ans = 1; while (expo){ if(expo & 1)(ans *= base) %= mod; expo >>= 1; base = base * base; base %= mod; } return ans % mod; } ll C(ll n, ll m){ if(n < m)return 0; ll x = 1, y = 1; if(m > n - m)m = n - m; rep(i ,0 , m - 1){ x = x * (n - i) % mod; y = y * (i + 1) % mod; } return x * quickPow(y, mod - 2) % mod; } ll lucas(ll n, ll m){ return !m || n == m ? 1 : C(n % mod, m % mod) * lucas(n / mod, m / mod) % mod; } int n, m, k; int a[limit]; ll fact[limit]; void calc(){ fact[0] = 1; rep(i, 1, limit - 1){ fact[i] = fact[i - 1] * i % mod; } } int top[limit]; ll ans = 1; int depth[limit]; ll tot; void dfs(int u, int dep, int tp){ depth[u] = dep; top[u] = tp; if(!top[a[u]]){ dfs(a[u], dep + 1, tp); }else if(top[u] == top[a[u]]){ (ans *= quickPow(2, dep - depth[a[u]] + 1) - 2) %= mod; tot += dep - depth[a[u]] + 1; } } void solve(){ cin>>n; rep(i,1,n){ cin>>a[i]; } rep(i, 1, n){ if(!top[i]){ dfs(i, 1, i); } } cout<<(((ans *= quickPow(2, n - tot))) %= mod)<<endl; }; int32_t main() { #ifdef LOCAL FOPEN; // FOUT; #endif FASTIO // int kase; // calc(); // cin>>kase; // while (kase--) solve(); cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << "s\n"; return 0; }
然后说到了基环树,好像没怎么涉及到这一块,所以做这种题就是直接两眼一抹黑,所以还是得看
然后看到了基环树定义,其实并不难理解,就是树根是一个环,然后其他地方和树的性质一样
记得刚开始学习算法竞赛的时候,看到过一篇NOI游记,说基环树要断边
那么来看看P5022旅行吧
所以这道题是问从某点出发,到能够到达的最大的地方,要求字典序最大的,题中很贴心地描述了回溯的过程
首先有向图很好搞,直接把边排序贪心就好了,但需要特殊处理的是基环树
题中某些情况(n == m)可以构成基环树,因为是基环树,我们需要从环上某一点切入,某一点改出,但我们不知道是从哪一点进行改出,所以我们枚举边,进行一个断边的操作,之后不断暴力比较字典序就好了
下周就学基环树好了
很奇妙
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include <bits/stdc++.h> using namespace std; constexpr int limit = (50000 + 5);//防止溢出 #define INF 0x3f3f3f3f #define inf 0x3f3f3f3f3f #define lowbit(i) i&(-i)//一步两步 #define EPS 1e-9 #define FASTIO ios::sync_with_stdio(false);cin.tie(0),cout.tie(0); #define ff(a) printf("%d\n",a ); #define pi(a, b) pair<a,b> #define rep(i, a, b) for(ll i = a; i <= b ; ++i) #define per(i, a, b) for(ll i = b ; i >= a ; --i) #define MOD 998244353 #define traverse(u) for(int i = head[u]; ~i ; i = edge[i].next) #define FOPEN freopen("C:\\Users\\tiany\\CLionProjects\\akioi\\data.txt", "rt", stdin) #define FOUT freopen("C:\\Users\\tiany\\CLionProjects\\akioi\\dabiao.txt", "wt", stdout) typedef long long ll; typedef unsigned long long ull; char buf[1 << 23], *p1 = buf, *p2 = buf, obuf[1 << 23], *O = obuf; inline ll read() { #define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++) ll sign = 1, x = 0; char s = getchar(); while (s > '9' || s < '0') { if (s == '-')sign = -1; s = getchar(); } while (s >= '0' && s <= '9') { x = (x << 3) + (x << 1) + s - '0'; s = getchar(); } return x * sign; #undef getchar }//快读 void print(ll x) { if (x / 10) print(x / 10); *O++ = x % 10 + '0'; } void write(ll x, char c = 't') { if (x < 0)putchar('-'), x = -x; print(x); if (!isalpha(c))*O++ = c; fwrite(obuf, O - obuf, 1, stdout); O = obuf; } constexpr ll mod = 1e9 + 7; ll quickPow(ll base, ll expo){ ll ans = 1; while (expo){ if(expo & 1)(ans *= base) %= mod; expo >>= 1; base = base * base; base %= mod; } return ans % mod; } ll C(ll n, ll m){ if(n < m)return 0; ll x = 1, y = 1; if(m > n - m)m = n - m; rep(i ,0 , m - 1){ x = x * (n - i) % mod; y = y * (i + 1) % mod; } return x * quickPow(y, mod - 2) % mod; } ll lucas(ll n, ll m){ return !m || n == m ? 1 : C(n % mod, m % mod) * lucas(n / mod, m / mod) % mod; } int n, m, k; int a[limit<<1]; ll fact[limit]; void calc(){ fact[0] = 1; rep(i, 1, limit - 1){ fact[i] = fact[i - 1] * i % mod; } } vector<int>g[limit], ans = {0}, temp = {0}; int vis[limit]; void add(int x, int y){ g[x].push_back(y); g[y].push_back(x); } void dfs(int u, int pre, int delu, int delv){ if(vis[u])return; vis[u] = 1; temp.push_back(u); for(auto v : g[u]){ if(v == pre or min(u, v) == min(delu, delv) and max(u, v) == max(delu, delv))continue; dfs(v, u, delu, delv); } } void solve(){ cin>>n>>m; rep(i,1,m){ cin>>a[i]>>a[i + m + 1]; add(a[i], a[i + m + 1]); } rep(i,1,n){ ranges::sort(g[i]); } if(n == m){ rep(i,1,m){ int u = a[i], v = a[i + m + 1]; temp.clear(); temp.push_back(0); memset(vis, 0, sizeof vis); dfs(1, -1, u, v); if(temp.size() < n + 1){ continue; } if(ans.size() == 1 or temp < ans){ ans.swap(temp); } } }else{ dfs(1, -1, -1, -1); ans.swap(temp); } for(auto && it : ans | views::drop(1)){ cout<<it<<' '; } }; int32_t main() { #ifdef LOCAL FOPEN; // FOUT; #endif FASTIO // int kase; // calc(); // cin>>kase; // while (kase--) solve(); cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << "s\n"; return 0; }
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