BZOJ 3653: 谈笑风生(DFS序+可持久化线段树)

首先嘛，还是太弱了，想了好久QAQ

然后，这道题么，明显就是求sigma（size[x]） (x是y的儿子且层树小于k) 然后就可以发现：把前n个节点按深度建可持久化线段树，就能用前缀和维护了

其实不难打= =

#include<cstdio>

#include<iostream>

#include<cstring>

#include<algorithm>

using namespace std;

#define maxn 300010

#define maxm 30000100

struct edges{

int to,next;

}edge[maxn*2];

int next[maxn],l,num;

int addedge(int x,int y){

edge[++l]=(edges){x,next[y]};next[y]=l;

edge[++l]=(edges){y,next[x]};next[x]=l;

return 0;

}

int id[maxn],d[maxn],s[maxn],e[maxn],b[maxn];

int dfs(int x,int y){

id[++num]=x;

d[x]=d[y]+1;

b[x]=num;

for (int i=next[x];i;i=edge[i].next){

if (edge[i].to!=y) {

dfs(edge[i].to,x);

s[x]+=s[edge[i].to]+1;

}

}

e[x]=num;

return 0;

}

struct node{

int lc,rc;long long s;

}t[maxm];

#define lc(x) t[x].lc

#define rc(x) t[x].rc

#define s(x) t[x].s

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

int build(int x,int l,int r){

if (l!=r) {

lc(x)=build(++num,l,mid);

rc(x)=build(++num,mid+1,r);

}

return x;

}

int ins(int x,int l,int r,int c,int z){

int y=++num;

if (l==r) {s(y)=s(x)+z;return y;}

lc(y)=lc(x);rc(y)=rc(x);

if (mid<c) rc(y)=ins(rc(x),mid+1,r,c,z);

else lc(y)=ins(lc(x),l,mid,c,z);

s(y)=s(lc(y))+s(rc(y));

return y;

}

long long sum(int x,int l,int r,int x1,int y1){

if (l>y1||r<x1) return 0;

if (x1<=l&&r<=y1) return s(x);

if (l==r) return s(x);

return sum(lc(x),l,mid,x1,y1)+sum(rc(x),mid+1,r,x1,y1);

}

int root[maxn];

int main(){

int n,q;

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

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

int x,y;

scanf("%d%d",&x,&y);

addedge(x,y);

}

dfs(1,0);

num=0;l=0;

root[0]=++num;

build(1,1,n);

for (int i=1;i<=n;i++)root[i]=ins(root[i-1],1,n,d[id[i]],s[id[i]]);

for (int i=1;i<=q;i++) {

int u,v;

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

printf("%lld\n",s[u]*1ll*min(v,d[u]-1)+sum(root[e[u]],1,n,d[u]+1,d[u]+v)-sum(root[b[u]],1,n,d[u]+1,d[u]+v));

}

return 0;

}