0.前言
若干年前会过一次,然后就不会了。
现在又会了。
1.虚树
存在这样一类问题,多次询问,每次给你 \(k\) 个特殊/关键点,要求有关这 \(k\) 个点的某种信息,并且题目保证了或隐含了 \(\sum k \le lim\),那么可以考虑使用虚树解决。
虚树,即将这 \(k\) 个点提出,并按照原树上的节点关系,建立一棵大小约为 \(2k\) 的树,可以在这棵树上求答案。
这里仅介绍二次排序 + LCA 的建树方法。
- 将关键点按照 dfs 序排序;
- 将排序后相邻两点的 LCA 加入,再按 dfs 序排序并去重;
- 连 \((LCA(ve_{i - 1},ve_i),ve_i)\) 的边。
对于 2:一般会将点 \(1\) 也加入,便于求答案。
因为按照 dfs 序排序了,所以相邻两点之间一定不会出现 dfs 序介于他们俩到 LCA 的路径上的点,正确性得证。
建树时间复杂度:\(\mathcal{O(\sum k \log n)}\)。
2.题目
2.1.P2495 [SDOI2011] 消耗战
板子题。
我们将虚树建出,接下来考虑树形 DP。
\(f_u\) 表示点 \(u\) 与其子树内所有关键点断开所需最小代价。
对于虚树上 \((u,v)\):若 \(v\) 是关键点,那么这条边必须断;否则判断是断这条边代价更小还是在 \(v\) 的子树中断若干边代价更小。
显然答案为 \(f_1\)。- #include <bits/stdc++.h>
- #define int long long
- #define ll long long
- #define ull unsigned long long
- #define db double
- #define ld long double
- #define rep(i,l,r) for (int i = (int)(l); i <= (int)(r); ++ i )
- #define rep1(i,l,r) for (int i = (int)(l); i >= (int)(r); -- i )
- #define il inline
- #define fst first
- #define snd second
- #define ptc putchar
- #define Yes ptc('Y'),ptc('e'),ptc('s'),puts("")
- #define No ptc('N'),ptc('o'),puts("")
- #define YES ptc('Y'),ptc('E'),ptc('S'),puts("")
- #define NO ptc('N'),ptc('O'),puts("")
- #define vi vector<int>
- #define pb emplace_back
- #define sz(x) (int)(x.size())
- #define all(x) x.begin(),x.end()
- #define me(a,x) memset(a,x,sizeof a)
- #define get(x) ((x - 1) / len + 1)
- #define debug() puts("------------")
- using namespace std;
- typedef pair<int,int> PII;
- typedef pair<int,PII> PIII;
- typedef pair<ll,ll> PLL;
- namespace szhqwq {
- template<class T> il void read(T &x) {
- x = 0; T f = 1; char ch = getchar();
- while (ch < '0' || ch > '9') { if (ch == '-') f = -1; ch = getchar(); }
- while (ch >= '0' && ch <= '9') { x = (x << 3) + (x << 1) + (ch ^ 48); ch = getchar(); }
- x *= f;
- }
- template<class T,class... Args> il void read(T &x,Args &...x_) { read(x); read(x_...); }
- template<class T> il void print(T x) {
- if (x < 0) ptc('-'), x = -x;
- if (x > 9) print(x / 10); ptc(x % 10 + '0');
- }
- template<class T,class T_> il void write(T x,T_ ch) { print(x); ptc(ch); }
- template<class T,class T_> il void chmax(T &x,T_ y) { x = x < (T)y ? (T)y : x; }
- template<class T,class T_> il void chmin(T &x,T_ y) { x = x > (T)y ? (T)y : x; }
- template<class T,class T_,class T__> il T qmi(T a,T_ b,T__ p) {
- T res = 1; while (b) {
- if (b & 1) res = res * a % p;
- a = a * a % p; b >>= 1;
- } return res;
- }
- template<class T> il T gcd(T a,T b) { if (!b) return a; return gcd(b,a % b); }
- template<class T,class T_> il void exgcd(T a, T b, T_ &x, T_ &y) {
- if (b == 0) { x = 1; y = 0; return; }
- exgcd(b,a % b,y,x); y -= a / b * x; return ;
- }
- template<class T,class T_> il T getinv(T x,T_ p) {
- T inv,y; exgcd(x,(T)p,inv,y);
- inv = (inv + p) % p; return inv;
- }
- } using namespace szhqwq;
- const int N = 5e5 + 10,inf = 1e9,mod = 998244353;
- const ull base = 131,base_ = 233;
- const ll inff = 1e18;
- const db eps = 1e-6;
- int n,m; vector<PII> G[N];
- int h[N],e[N],ne[N],w[N],idx,dfn[N],tot;
- int d[N],fa[N][20],dis[N][20],f[N];
- bool st[N];
- il void add(int a,int b,int c) {
- e[idx] = b;
- w[idx] = c;
- ne[idx] = h[a];
- h[a] = idx ++;
- return ;
- }
- il void dfs(int u,int faa,int val) {
- dfn[u] = ++ tot;
- fa[u][0] = faa; dis[u][0] = val;
- d[u] = d[faa] + 1;
- rep(i,1,18)
- fa[u][i] = fa[fa[u][i - 1]][i - 1],
- dis[u][i] = min(dis[u][i - 1],dis[fa[u][i - 1]][i - 1]);
- for (int i = h[u]; ~i; i = ne[i]) {
- int j = e[i];
- if (j == faa) continue;
- dfs(j,u,w[i]);
- }
- return ;
- }
- il int LCA(int x,int y) {
- if (d[x] < d[y]) swap(x,y);
- rep1(i,18,0) if (d[fa[x][i]] >= d[y]) x = fa[x][i];
- if (x == y) return x;
- rep1(i,18,0) if (fa[x][i] != fa[y][i]) x = fa[x][i],y = fa[y][i];
- return fa[x][0];
- }
- il int calc(int x,int p) {
- int ret = inf;
- rep1(i,18,0) if (d[fa[x][i]] >= d[p])
- chmin(ret,dis[x][i]),x = fa[x][i];
- return ret;
- }
- il void dfss(int u,int faa) {
- f[u] = 0;
- for (auto x : G[u]) {
- int v = x.fst,w = x.snd;
- if (v == faa) continue;
- dfss(v,u);
- if (st[v]) f[u] += w;
- else f[u] += min(f[v],w);
- }
- return ;
- }
- il void solve() {
- //------------code------------
- me(h,-1); me(dis,0x3f);
- read(n);
- rep(i,1,n - 1) {
- int a,b,c; read(a,b,c);
- add(a,b,c); add(b,a,c);
- }
- dfs(1,0,inf);
- read(m);
- while (m -- ) {
- int k; read(k);
- vi v,ve;
- rep(i,1,k) {
- int x; read(x);
- v.pb(x);
- }
- sort(all(v),[](int x,int y){ return dfn[x] < dfn[y]; });
- ve.pb(1); ve.pb(v[0]);
- rep(i,1,k - 1)
- ve.pb(LCA(v[i - 1],v[i])),
- ve.pb(v[i]);
- sort(all(ve),[](int x,int y){ return dfn[x] < dfn[y]; });
- ve.erase(unique(all(ve)),ve.end());
- rep(i,0,sz(ve) - 1) st[ve[i]] = 0,G[ve[i]].clear();
- for (auto x : v) st[x] = 1;
- rep(i,1,sz(ve) - 1) {
- int lca = LCA(ve[i - 1],ve[i]);
- G[lca].pb(ve[i],calc(ve[i],lca));
- }
- dfss(1,0); write(f[1],'\n');
- }
- return ;
- }
- il void init() {
- return ;
- }
- signed main() {
- // init();
- int _ = 1;
- // read(_);
- while (_ -- ) solve();
- return 0;
- }
复制代码 2.2.CF613D Kingdom and its Cities
同样建出虚树。要使所有关键点两两不连通,因为点权相同,可以直接贪心。
先判掉无解的情况。
若当前点 \(u\) 是关键点且其子树中存在其他关键点,那么子树中有多少关键点答案就加多少;
否则看子树中有多少关键点,若有 \(> 1\) 个,那么占领 \(u\) 点一定不劣;如果仅有 \(1\) 个,考虑放到父亲节点及上面进行处理一定更优。- #include <bits/stdc++.h>
- // #define int long long
- #define ll long long
- #define ull unsigned long long
- #define db double
- #define ld long double
- #define rep(i,l,r) for (int i = (int)(l); i <= (int)(r); ++ i )
- #define rep1(i,l,r) for (int i = (int)(l); i >= (int)(r); -- i )
- #define il inline
- #define fst first
- #define snd second
- #define ptc putchar
- #define Yes ptc('Y'),ptc('e'),ptc('s'),puts("")
- #define No ptc('N'),ptc('o'),puts("")
- #define YES ptc('Y'),ptc('E'),ptc('S'),puts("")
- #define NO ptc('N'),ptc('O'),puts("")
- #define vi vector<int>
- #define pb emplace_back
- #define sz(x) (int)(x.size())
- #define all(x) x.begin(),x.end()
- #define me(a,x) memset(a,x,sizeof a)
- #define get(x) ((x - 1) / len + 1)
- #define debug() puts("------------")
- using namespace std;
- typedef pair<int,int> PII;
- typedef pair<int,PII> PIII;
- typedef pair<ll,ll> PLL;
- namespace szhqwq {
- template<class T> il void read(T &x) {
- x = 0; T f = 1; char ch = getchar();
- while (ch < '0' || ch > '9') { if (ch == '-') f = -1; ch = getchar(); }
- while (ch >= '0' && ch <= '9') { x = (x << 3) + (x << 1) + (ch ^ 48); ch = getchar(); }
- x *= f;
- }
- template<class T,class... Args> il void read(T &x,Args &...x_) { read(x); read(x_...); }
- template<class T> il void print(T x) {
- if (x < 0) ptc('-'), x = -x;
- if (x > 9) print(x / 10); ptc(x % 10 + '0');
- }
- template<class T,class T_> il void write(T x,T_ ch) { print(x); ptc(ch); }
- template<class T,class T_> il void chmax(T &x,T_ y) { x = x < (T)y ? (T)y : x; }
- template<class T,class T_> il void chmin(T &x,T_ y) { x = x > (T)y ? (T)y : x; }
- template<class T,class T_,class T__> il T qmi(T a,T_ b,T__ p) {
- T res = 1; while (b) {
- if (b & 1) res = res * a % p;
- a = a * a % p; b >>= 1;
- } return res;
- }
- template<class T> il T gcd(T a,T b) { if (!b) return a; return gcd(b,a % b); }
- template<class T,class T_> il void exgcd(T a, T b, T_ &x, T_ &y) {
- if (b == 0) { x = 1; y = 0; return; }
- exgcd(b,a % b,y,x); y -= a / b * x; return ;
- }
- template<class T,class T_> il T getinv(T x,T_ p) {
- T inv,y; exgcd(x,(T)p,inv,y);
- inv = (inv + p) % p; return inv;
- }
- } using namespace szhqwq;
- const int N = 2e5 + 10,inf = 1e9,mod = 998244353;
- const ull base = 131,base_ = 233;
- const ll inff = 1e18;
- const db eps = 1e-6;
- int n,q,id[N],cnt,fa[N][20];
- int h[N],e[N],ne[N],idx,d[N];
- bool vis[N];
- vi G[N];
- il void add(int a,int b) {
- e[idx] = b;
- ne[idx] = h[a];
- h[a] = idx ++;
- return ;
- }
- il void dfs(int u,int f) {
- fa[u][0] = f;
- d[u] = d[f] + 1;
- rep(i,1,18) fa[u][i] = fa[fa[u][i - 1]][i - 1];
- id[u] = ++ cnt;
- for (int i = h[u]; ~i; i = ne[i]) {
- int j = e[i];
- if (j == f) continue;
- dfs(j,u);
- }
- return ;
- }
- il int LCA(int x,int y) {
- if (d[x] < d[y]) swap(x,y);
- rep1(i,18,0) if (d[fa[x][i]] >= d[y]) x = fa[x][i];
- if (x == y) return x;
- rep1(i,18,0) if (fa[x][i] != fa[y][i]) x = fa[x][i],y = fa[y][i];
- return fa[x][0];
- }
- int ret = 0,st[N];
- il void calcans(int u) {
- if (vis[u]) st[u] = 1;
- int tot = 0;
- for (auto v : G[u]) {
- calcans(v);
- tot += st[v];
- }
- if (vis[u] && tot) ret += tot;
- else if (tot == 1) st[u] = 1;
- else if (tot > 1) ++ ret;
- return ;
- }
- il void solve() {
- //------------code------------
- read(n); me(h,-1);
- rep(i,1,n - 1) {
- int a,b; read(a,b);
- add(a,b); add(b,a);
- }
- dfs(1,0);
- read(q);
- while (q -- ) {
- int k; read(k);
- vi v,ve;
- rep(i,1,k) {
- int x; read(x);
- v.pb(x);
- }
- sort(all(v),[](int x,int y){ return id[x] < id[y]; });
- ve.pb(1); ve.pb(v[0]);
- rep(i,1,sz(v) - 1)
- ve.pb(LCA(v[i - 1],v[i])),
- ve.pb(v[i]);
- sort(all(ve),[](int x,int y){ return id[x] < id[y]; });
- ve.erase(unique(all(ve)),ve.end());
- for (auto x : ve) G[x].clear(),vis[x] = vis[fa[x][0]] = 0,st[x] = 0;
- for (auto x : v) vis[x] = 1;
- bool fl = 1;
- for (auto x : v) if (vis[fa[x][0]]) { fl = 0; break; }
- if (!fl) { puts("-1"); continue; }
- rep(i,1,sz(ve) - 1) G[LCA(ve[i - 1],ve[i])].pb(ve[i]);
- ret = 0;
- calcans(1);
- write(ret,'\n');
- }
- return ;
- }
- il void init() {
- return ;
- }
- signed main() {
- // init();
- int _ = 1;
- // read(_);
- while (_ -- ) solve();
- return 0;
- }
复制代码 2.3.P3233 [HNOI2014] 世界树
建虚树,考虑 up and down DP。
\(f_u = (dist,id)\),分别表示最短距离及编号。
分别向上向下更新信息。
考虑最后计算答案怎么做。
对于虚树上 \((u,v)\),令 \(p\) 为 \(u\) 在原树上和 \(v\) 那条链上的儿子。
如果两点所属的关键点相同,则 \(cnt_{id} \gets siz_p - siz_v\)。
若不同,则倍增跳到分界点,即上半部分所属点与 \(u\) 相同,下半部分所属点与 \(v\) 相同的点 \(mid\),\(mid\) 自己和 \(y\) 相同。
\(cnt_{id_u} \gets siz_p - siz_{mid},cnt_{id_v} \gets siz_{mid} - siz_v\)。
注意到 \(u\) 某些子树中可能不存在关键点,那么这些子树中的点显然所属与 \(u\) 一致,处理一下即可。- #include <bits/stdc++.h>
- // #define int long long
- #define ll long long
- #define ull unsigned long long
- #define db double
- #define ld long double
- #define rep(i,l,r) for (int i = (int)(l); i <= (int)(r); ++ i )
- #define rep1(i,l,r) for (int i = (int)(l); i >= (int)(r); -- i )
- #define il inline
- #define fst first
- #define snd second
- #define ptc putchar
- #define Yes ptc('Y'),ptc('e'),ptc('s'),puts("")
- #define No ptc('N'),ptc('o'),puts("")
- #define YES ptc('Y'),ptc('E'),ptc('S'),puts("")
- #define NO ptc('N'),ptc('O'),puts("")
- #define vi vector<int>
- #define pb emplace_back
- #define sz(x) (int)(x.size())
- #define all(x) x.begin(),x.end()
- #define me(a,x) memset(a,x,sizeof a)
- #define get(x) ((x - 1) / len + 1)
- #define debug() puts("------------")
- using namespace std;
- typedef pair<int,int> PII;
- typedef pair<int,PII> PIII;
- typedef pair<ll,ll> PLL;
- namespace szhqwq {
- template<class T> il void read(T &x) {
- x = 0; T f = 1; char ch = getchar();
- while (ch < '0' || ch > '9') { if (ch == '-') f = -1; ch = getchar(); }
- while (ch >= '0' && ch <= '9') { x = (x << 3) + (x << 1) + (ch ^ 48); ch = getchar(); }
- x *= f;
- }
- template<class T,class... Args> il void read(T &x,Args &...x_) { read(x); read(x_...); }
- template<class T> il void print(T x) {
- if (x < 0) ptc('-'), x = -x;
- if (x > 9) print(x / 10); ptc(x % 10 + '0');
- }
- template<class T,class T_> il void write(T x,T_ ch) { print(x); ptc(ch); }
- template<class T,class T_> il void chmax(T &x,T_ y) { x = x < (T)y ? (T)y : x; }
- template<class T,class T_> il void chmin(T &x,T_ y) { x = x > (T)y ? (T)y : x; }
- template<class T,class T_,class T__> il T qmi(T a,T_ b,T__ p) {
- T res = 1; while (b) {
- if (b & 1) res = res * a % p;
- a = a * a % p; b >>= 1;
- } return res;
- }
- template<class T> il T gcd(T a,T b) { if (!b) return a; return gcd(b,a % b); }
- template<class T,class T_> il void exgcd(T a, T b, T_ &x, T_ &y) {
- if (b == 0) { x = 1; y = 0; return; }
- exgcd(b,a % b,y,x); y -= a / b * x; return ;
- }
- template<class T,class T_> il T getinv(T x,T_ p) {
- T inv,y; exgcd(x,(T)p,inv,y);
- inv = (inv + p) % p; return inv;
- }
- } using namespace szhqwq;
- const int N = 3e5 + 10,inf = 1e9,mod = 998244353;
- const ull base = 131,base_ = 233;
- const ll inff = 1e18;
- const db eps = 1e-6;
- int n,q,d[N],fa[N][20],siz[N];
- int h[N],e[N << 1],ne[N << 1],idx;
- PII f[N]; int id[N],cnt;
- vi G[N]; bool vis[N];
- int ret[N];
- il void add(int a,int b) {
- e[idx] = b;
- ne[idx] = h[a];
- h[a] = idx ++;
- return ;
- }
- il void dfs(int u,int f) {
- d[u] = d[f] + 1;
- fa[u][0] = f; id[u] = ++ cnt;
- rep(i,1,18) fa[u][i] = fa[fa[u][i - 1]][i - 1];
- siz[u] = 1;
- for (int i = h[u]; ~i; i = ne[i]) {
- int j = e[i];
- if (j == f) continue;
- dfs(j,u);
- siz[u] += siz[j];
- }
- return ;
- }
- il int LCA(int x,int y) {
- if (d[x] < d[y]) swap(x,y);
- rep1(i,18,0) if (d[fa[x][i]] >= d[y]) x = fa[x][i];
- if (x == y) return x;
- rep1(i,18,0) if (fa[x][i] != fa[y][i]) x = fa[x][i],y = fa[y][i];
- return fa[x][0];
- }
- il int getdis(int x,int y) {
- return d[x] + d[y] - 2 * d[LCA(x,y)];
- }
- il void dfs1(int u) {
- f[u] = {inf,0};
- if (vis[u]) f[u] = {0,u};
- for (auto v : G[u]) {
- dfs1(v);
- if (f[v].snd) {
- int val = getdis(u,f[v].snd);
- if (val < f[u].fst) f[u] = {val,f[v].snd};
- else if (val == f[u].fst) chmin(f[u].snd,f[v].snd);
- }
- }
- return ;
- }
- il void dfs2(int u) {
- for (auto v : G[u]) {
- if (f[u].snd) {
- int val = getdis(v,f[u].snd);
- if (val < f[v].fst) f[v] = {val,f[u].snd};
- else if (val == f[v].fst) chmin(f[v].snd,f[u].snd);
- }
- dfs2(v);
- }
- return ;
- }
- il void calcans(int u) {
- int val = siz[u];
- for (auto v : G[u]) {
- int p = v;
- rep1(i,18,0) if (d[fa[p][i]] > d[u]) p = fa[p][i];
- val -= siz[p];
- if (f[u].snd == f[v].snd) ret[f[u].snd] += siz[p] - siz[v];
- else {
- int mid = v;
- rep1(i,18,0) {
- int dis = getdis(f[u].snd,fa[mid][i]),diss = getdis(f[v].snd,fa[mid][i]);
- if (dis > diss || dis == diss && f[v].snd < f[u].snd) mid = fa[mid][i];
- }
- ret[f[u].snd] += siz[p] - siz[mid]; ret[f[v].snd] += siz[mid] - siz[v];
- }
- calcans(v);
- }
- ret[f[u].snd] += val;
- return ;
- }
- il void solve() {
- //------------code------------
- read(n); me(h,-1);
- rep(i,1,n - 1) {
- int a,b; read(a,b);
- add(a,b); add(b,a);
- }
- dfs(1,0);
- read(q);
- while (q -- ) {
- int m; read(m);
- vi v,ve,vec;
- rep(i,1,m) {
- int x; read(x);
- v.pb(x); vec.pb(x);
- ret[x] = 0;
- }
- sort(all(v),[](int x,int y){ return id[x] < id[y]; });
- ve.pb(1); ve.pb(v[0]);
- rep(i,1,sz(v) - 1)
- ve.pb(LCA(v[i - 1],v[i])),
- ve.pb(v[i]);
- sort(all(ve),[](int x,int y){ return id[x] < id[y]; });
- ve.erase(unique(all(ve)),ve.end());
- for (auto x : ve) G[x].clear(),vis[x] = 0;
- for (auto x : v) vis[x] = 1;
- rep(i,1,sz(ve) - 1) G[LCA(ve[i - 1],ve[i])].pb(ve[i]);
- dfs1(1); dfs2(1); calcans(1);
- for (auto x : vec) write(ret[x],' ');
- puts("");
- }
- return ;
- }
- il void init() {
- return ;
- }
- signed main() {
- // init();
- int _ = 1;
- // read(_);
- while (_ -- ) solve();
- return 0;
- }
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