BZOJ 4538: [Hnoi2016]网络 [整体二分]

4538: [Hnoi2016]网络

题意：一棵树，支持添加一条u到v权值为k的路径，删除之前的一条路径，询问不经过点x的路径的最大权值

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考虑二分

整体二分最大权值，如果\(k \in [mid+1,r]\)中的路径有不经过x的，那么这个询问的答案在\([mid+1,r]\)中

链修改，点查询\(\rightarrow\)点修改，子树查询，方法是\(u,v +1\ ;\ lca,fa[lca] -1\)

用树状数组就可以完成

这里的整体二分不需要对每个询问保存当前贡献，因为每次只需要考虑一段的贡献

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
#define fir first
#define sec second
const int N=2e5+5;
inline int read(){
    char c=getchar();int x=0,f=1;
    while(c<'0'||c>'9'){if(c=='-')f=-1; c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0'; c=getchar();}
    return x*f;
}

int n, Q, id[N], mp[N], t1[N], t2[N], ans[N];
struct meow{int op, u, v, k, t, x;} q[N];

struct edge{int v, ne;}e[N-1];
int cnt=1, h[N];
inline void ins(int u, int v) {
	e[++cnt]=(edge){v, h[u]}; h[u]=cnt;
	e[++cnt]=(edge){u, h[v]}; h[v]=cnt;
}
namespace tr {
	pair<int, int> dfn[N];
	int fa[N], deep[N], dfc, tot, pos[N], f[N<<1][18];
	void dfs(int u) {
		dfn[u].fir = ++dfc;
		f[++tot][0] = u;
		pos[u] = tot;
		for(int i=h[u];i;i=e[i].ne)
			if(e[i].v != fa[u]) {
				fa[e[i].v] = u;
				deep[e[i].v] = deep[u]+1;
				dfs(e[i].v);
				f[++tot][0] = u;
			}
		dfn[u].sec = dfc;
	}
	int log[N];
	inline int min(int x, int y) {return deep[x] < deep[y] ? x : y;}
	void init() {
		dfs(1);
		//for(int i=1; i<=tot; i++) printf("%d ", f[i][0]); puts("");
		for(int j=1; j<=17; j++)
			for(int i=1; i+(1<<j)-1<=tot; i++)
				f[i][j] = min(f[i][j-1], f[i+(1<<(j-1))][j-1]);// printf("f %d %d  %d\n",i,j,f[i][j]);

		log[1]=0; for(int i=2; i<=tot; i++) log[i] = log[i>>1]+1;
	}
	inline int lca(int x, int y) {
		x = pos[x], y = pos[y]; if(x>y) swap(x, y);
		int t = log[y-x+1];
		return min(f[x][t], f[y-(1<<t)+1][t]);
	}
} using tr::lca; using tr::fa; using tr::dfn;

namespace bit {
	int c[N], T, t[N];
	inline void ini() {T++;}
	inline void add(int p, int v) {
		for(; p<=n; p+=p&-p) {
			if(t[p] == T) c[p] += v;
			else t[p] = T, c[p] = v;
		}
	}
	inline int sum(int p) {
		int ans=0;
		for(; p; p-=p&-p) if(t[p] == T) ans += c[p];
		return ans;
	}
	inline int sum(int l, int r) {return sum(r) - sum(l-1);}
	inline void cha(int id, int flag) {
		int u=q[id].u, v=q[id].v, p=lca(u, v); //printf("uvp %d %d %d\n", u, v, p);
		add(dfn[u].fir, flag); add(dfn[v].fir, flag);
		add(dfn[p].fir, -flag); if(fa[p]) add(dfn[ fa[p] ].fir, -flag);
	}
} using bit::sum; using bit::cha;

void cdq(int l, int r, int ql, int qr) { //printf("cdq [%d, %d]  %d %d\n",l, r, ql, qr);
	if(l==r) {
		for(int i=ql; i<=qr; i++) if(q[id[i]].op == 3) ans[ id[i] ] = l;
		return;
	}

	int mid = (l+r)>>1, p1=0, p2=0; //printf("mid %d\n", mid);
	bit::ini();
	int now=0;
	for(int i=ql; i<=qr; i++) {
		int _=i; i=id[i]; //printf("i-------- %d %d  %d\n",_,i, q[i].op);
		if(q[i].op == 1) { //printf("k %d\n", q[i].k);
			if(q[i].k > mid) cha(i, 1), now++, t2[++p2] = i;
			else t1[++p1] = i;
		} else if(q[i].op == 2) {
			if(q[ q[i].t ].k > mid) cha(q[i].t, -1), now--, t2[++p2] = i;
			else t1[++p1] = i;
		} else {
			int x = q[i].x, cnt = sum(dfn[x].fir, dfn[x].sec); //printf("hi que %d  %d %d\n", x, cnt, now);
			if(cnt < now) t2[++p2] = i;
			else t1[++p1] = i;
		}
		i=_;
	}
	for(int i=1; i<=p1; i++) id[ql+i-1] = t1[i];
	for(int i=1; i<=p2; i++) id[ql+p1+i-1] = t2[i];
	cdq(l, mid, ql, ql+p1-1); cdq(mid+1, r, ql+p1, qr);
}

int main() {
	//freopen("in", "r", stdin);
	freopen("network_tenderRun.in", "r", stdin);
	freopen("network_tenderRun.out", "w", stdout);
	n=read(); Q=read();
	for(int i=1; i<n; i++) ins(read(), read());
	for(int i=1; i<=Q; i++) {
		q[i].op = read()+1;
		if(q[i].op == 1) q[i].u=read(), q[i].v=read(), mp[++mp[0]] = q[i].k=read();
		else if(q[i].op == 2) q[i].t=read();
		else q[i].x=read();
		id[i]=i;
	}
	sort(mp+1, mp+mp[0]+1); mp[0] = unique(mp+1, mp+mp[0]+1) - mp - 1;
	for(int i=1; i<=Q; i++) if(q[i].op == 1) q[i].k = lower_bound(mp+1, mp+mp[0]+1, q[i].k) - mp;// printf("k %d  %d\n", i, q[i].k);
	tr::init();
	//for(int i=1; i<=n; i++) for(int j=1; j<=n; j++) printf("lca %d %d  %d\n",i,j,lca(i, j));
	cdq(0, mp[0], 1, Q);
	for(int i=1; i<=Q; i++) if(q[i].op == 3) printf("%d\n", ans[i] ? mp[ans[i]] : -1);
}