bzoj 1267 Kth Number I (点分治,堆)

超级钢琴的树上版本, 类似做法即可, 只不过区间转为dfs序了, 用点分求一下, 复杂度$O(nlog^2n)$

#include <iostream>

#include <algorithm>

#include <cstdio>

#include <math.h>

#include <set>

#include <map>

#include <queue>

#include <string>

#include <string.h>

#include <bitset>

#define REP(i,a,n) for(int i=a;i<=n;++i)

#define PER(i,a,n) for(int i=n;i>=a;--i)

#define hr putchar(10)

#define pb push_back

#define lc (o<<1)

#define rc (lc|1)

#define mid ((l+r)>>1)

#define ls lc,l,mid

#define rs rc,mid+1,r

#define x first

#define y second

#define io std::ios::sync_with_stdio(false)

#define endl '\n'

using namespace std;

typedef long long ll;

typedef pair<int,int> pii;

const int P = 1e9+7, INF = 0x3f3f3f3f;

ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}

ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}

ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}

//head

const int N = 1e6+10;

int n, m, sum, rt;

struct _ {int to,w;};

vector<_> g[N];

int mx[N], sz[N], vis[N];

#define go for (int i=0,y,w; i<g[x].size()?w=g[x][i].w,y=g[x][i].to:0; ++i)

void getrt(int x, int fa) {

	mx[x]=0,sz[x]=1;

	go if (!vis[y]&&y!=fa) {

		getrt(y,x),sz[x]+=sz[y];

		mx[x]=max(mx[x],sz[y]);

	}

	mx[x]=max(mx[x],sum-sz[x]);

	if (mx[rt]>mx[x]) rt=x;

}

int s[N], L, R;

pii a[N];

void dfs(int x, int f, int d, int rt) {

	s[++*s]=d, a[*s]=pii(L,R);

	go if (!vis[y]&&y!=f) dfs(y,x,d+w,rt);

}

void solve(int x) {

	vis[x]=1,s[++*s]=0,L=R=*s;

	go if (!vis[y]) dfs(y,x,w,x),R=*s;

	go if (!vis[y]) {

		mx[rt=0]=n,sum=sz[y];

		getrt(y,0),solve(rt);

	}

}

int Log[N], f[N][20];

void init(int n) {

	Log[0]=-1;

	REP(i,1,n) f[i][0]=i,Log[i]=Log[i>>1]+1;

	for (int j=1; (1<<j)<=n; ++j) {

		for (int i=0; i+(1<<j)-1<=n; ++i) {

			int x=f[i][j-1], y=f[i+(1<<(j-1))][j-1];

			f[i][j]=s[x]>s[y]?x:y;

		}

	}

}

int RMQ(int l, int r) {

	int k=Log[r-l+1];

	int x=f[l][k],y=f[r-(1<<k)+1][k];

	return s[x]>s[y]?x:y;

}

struct node {

	int l,r,pos,opt,v;

	node () {}

	node (int l, int r, int pos) : l(l),r(r),pos(pos) {

		opt = RMQ(l,r);

		v = s[pos]+s[opt];

	}

	bool operator < (const node & rhs) const {

		return v < rhs.v;

	}

};

priority_queue<node> q;

int main() {

	scanf("%d%d", &n, &m);

	REP(i,2,n) {

		int u, v, w;

		scanf("%d%d%d", &u, &v, &w);

		g[u].pb({v,w}),g[v].pb({u,w});

	}

	sum=mx[0]=n,getrt(1,0),solve(rt);

	init(*s);

	REP(i,1,*s) if (a[i].x) q.push(node(a[i].x,a[i].y,i));

	REP(i,1,m) {

		node t = q.top();

		printf("%d\n", t.v);q.pop();

		if (t.opt!=t.l) q.push(node(t.l,t.opt-1,t.pos));

		if (t.opt!=t.r) q.push(node(t.opt+1,t.r,t.pos));

	}

}