题解 选择

传送门

• 当出现形如「每个结点最多与 \(k\) 条边相连」且 \(k\) 很小的时候，可以考虑树形DP+状压合并子树

基于一个挺套路的结论：

• 就题论题，这个题合并子树如果是状压枚举选哪些边的话只能过 \(n=5\)
  但如果压选择了哪些与根节点直接相连的边被选，对于每个状态枚举所有路径，如果两个端点所在子树都没被选过就转移则可以过 \(n=15\) 或更高

Code:

#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define N 1010
#define ll long long
#define reg register int
#define fir first
#define sec second
#define make make_pair
#define pb push_back
//#define int long long 

char buf[1<<21], *p1=buf, *p2=buf;
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf, 1, 1<<21, stdin)), p1==p2?EOF:*p1++)
inline int read() {
	int ans=0, f=1; char c=getchar();
	while (!isdigit(c)) {if (c=='-') f=-f; c=getchar();}
	while (isdigit(c)) {ans=(ans<<3)+(ans<<1)+(c^48); c=getchar();}
	return ans*f;
}

int n, m;
int head[N], size;
struct edge{int to, next, id;}e[N<<1];
inline void add(int s, int t, int i) {e[++size].to=t; e[size].id=i; e[size].next=head[s]; head[s]=size;}

namespace force{
	pair<int, int> q[N];
	bitset<1500> lim[10000];
	bool dfs(int u, int to, int fa, bitset<1500>& dat) {
		if (u==to) return 1;
		for (int i=head[u],v; ~i; i=e[i].next) {
			v = e[i].to;
			if (v!=fa && dfs(v, to, u, dat)) {dat[e[i].id]=1; return 1;}
		}
		return 0;
	}
	void solve() {
		for (int i=1,u,v; i<n; ++i) {
			u=read(); v=read();
			add(u, v, i); add(v, u, i);
		}
		m=read();
		for (int i=1; i<=m; ++i) {
			q[i].fir=read(); q[i].sec=read();
			dfs(q[i].fir, q[i].sec, 0, lim[i]);
		}
		int lim2=1<<m, ans=0;
		bitset<1500> used;
		for (int s=1,s2,cnt; s<lim2; ++s) {
			s2=s; cnt=0;
			do {++cnt; s2&=s2-1;} while (s2);
			if (cnt<=ans) goto jump;
			used.reset();
			for (int i=0; i<m; ++i) if (s&(1<<i)) {
				if ((used&lim[i+1]).any()) goto jump;
				else used|=lim[i+1];
			}
			ans=cnt;
			jump: ;
		}
		printf("%d\n", ans);
		exit(0);
	}
}

namespace task1{
	int ans, lst;
	struct line{int l, r; inline void build(int a, int b) {l=a; r=b;}}p[N];
	inline bool operator < (line a, line b) {return a.r==b.r?a.l<b.l:a.r<b.r;}
	void solve() {
		for (int i=1,u,v; i<n; ++i) {
			u=read(); v=read();
			add(u, v, i); add(v, u, i);
		}
		m=read();
		for (int i=1,u,v; i<=m; ++i) {
			u=read(); v=read();
			if (u>v) swap(u, v);
			p[i].build(u, v);
		}
		sort(p+1, p+m+1);
		++ans, lst=p[1].r;
		for (int i=2; i<=m; ++i) {
			if (p[i].l>=lst) ++ans, lst=p[i].r;
		}
		printf("%d\n", ans);
		exit(0);
	}
}

namespace task{
	int dp[N], rk[N], f[1<<11], top, e2[N];
	vector<int> e[N], son[N];
	pair<int, int> path[N*N];
	bool q[N][N], lock[12];
	void dfs(int u, int fa) {
		// cout<<"dfs: "<<u<<' '<<fa<<endl;
		son[u].pb(u);
		for (auto v:e[u]) if (v!=fa) dfs(v, u);
		int tot=0, top=0;
		memset(lock, 0, sizeof(lock));
		for (auto v:e[u]) if (v!=fa) {
			rk[v]=tot; e2[tot++]=v;
			dp[u]+=dp[v];
			for (auto i:son[v]) {
				// cout<<"i: "<<i<<endl;
				if (q[i][u]) {
					lock[tot-1]=1;
					break;
				}
				for (int j=0; j<tot-1; ++j)
					for (auto k:son[e2[j]]) {
						// cout<<"k: "<<k<<endl;
						if (q[i][k]) {
							path[++top]=make(tot-1, j);
							break;
						}
					}
			}
		}
		// cout<<"top: "<<top<<endl;
		// cout<<"tot: "<<tot<<endl;
		int lim=1<<tot, must=lim-1, maxn=0;
		memset(f, 0, sizeof(f));
		for (int s=0; s<lim; ++s) {
			maxn=max(maxn, f[s]);
			for (int i=0; i<tot; ++i) if (lock[i] && !(s&(1<<i))) f[s|(1<<i)]=max(f[s|(1<<i)], f[s]+1); //, cout<<1<<endl;
			for (int i=1; i<=top; ++i) {
				if (!(s&(1<<path[i].fir)) && !(s&(1<<path[i].sec)))
					f[s|(1<<path[i].fir)|(1<<path[i].sec)] = max(f[s|(1<<path[i].fir)|(1<<path[i].sec)], f[s]+1); //, cout<<2<<' '<<bitset<5>(s|(1<<path[i].fir)|(1<<path[i].sec))<<endl;
			}
		}
		if (maxn) {for (int s=1; s<lim; ++s) if (f[s]==maxn) must&=s;}
		else must=0;
		// cout<<"u: "<<u<<endl;
		// cout<<"maxn: "<<maxn<<endl;
		// cout<<"must: "<<must<<endl;
		dp[u]+=maxn;
		// cout<<"son: "; for (auto it:son[u]) cout<<it<<' '; cout<<endl;
		for (int i=0; i<tot; ++i) if (!(must&(1<<i))) {
			// cout<<"add: "<<i<<' '<<e2[i]<<endl;
			for (auto j:son[e2[i]]) son[u].pb(j);
		}
		// cout<<"son: "; for (auto it:son[u]) cout<<it<<' '; cout<<endl;
	}
	void solve() {
		for (int i=1,u,v; i<n; ++i) {
			u=read(); v=read();
			e[u].pb(v); e[v].pb(u);
		}
		m=read();
		for (int i=1,u,v; i<=m; ++i) {
			u=read(); v=read();
			q[u][v]=q[v][u]=1;
		}
		dfs(1, 0);
		printf("%d\n", dp[1]);
		exit(0);
	}
}

signed main()
{
	freopen("select.in", "r", stdin);
	freopen("select.out", "w", stdout);
	
	memset(head, -1, sizeof(head));
	n=read();
	// if (n<=5) force::solve();
	// else task1::solve();
	task::solve();

	return 0;
}