[Codeforces Round #447] 894D

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Ralph And His Tour in Binary Country

time limit per test2.5 seconds
memory limit per test512 megabytes
inputstandard input
outputstandard output

Ralph is in the Binary Country. The Binary Country consists of n cities and (n - 1) bidirectional roads connecting the cities. The roads are numbered from 1 to (n - 1), the i-th road connects the city labeled (here ⌊ x⌋ denotes the x rounded down to the nearest integer) and the city labeled (i + 1), and the length of the i-th road is Li.

Now Ralph gives you m queries. In each query he tells you some city Ai and an integer Hi. He wants to make some tours starting from this city. He can choose any city in the Binary Country (including Ai) as the terminal city for a tour. He gains happiness (Hi - L) during a tour, where L is the distance between the city Ai and the terminal city.

Ralph is interested in tours from Ai in which he can gain positive happiness. For each query, compute the sum of happiness gains for all such tours.

Ralph will never take the same tour twice or more (in one query), he will never pass the same city twice or more in one tour.

Input
The first line contains two integers n and m (1 ≤ n ≤ 106, 1 ≤ m ≤ 105).

(n - 1) lines follow, each line contains one integer Li (1 ≤ Li ≤ 105), which denotes the length of the i-th road.

m lines follow, each line contains two integers Ai and Hi (1 ≤ Ai ≤ n, 0 ≤ Hi ≤ 107).

Output
Print m lines, on the i-th line print one integer — the answer for the i-th query.

Examples
input
2 2
5
1 8
2 4
output
11
4
input
6 4
2
1
1
3
2
2 4
1 3
3 2
1 7
output
11
6
3
28

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给你一个树，边的连接方式是i与i/2连一条无向边，长度为l[i]。q个询问，每个询问给出一个点a和一个值H，求以a为起点到任意点为终点获得（happiness=H-路径长度）不为负数的所有happiness之和。

看了题解才弄懂的，每一个点递增有序地储存以它为根节点，所有子节点（包括自己）到它的距离并处理出前缀和。所需总空间为O(nlog(n))，然后对于每个询问，从询问点不断往上爬（编号/2就是了）并且减去边长,在每一层时，二分查询左右子树lch和rch（不是刚爬上来的那个子树）中满足条件（总路径小于H）的最大编号，然后就是直接利用前缀和贡献答案就是了。时间复杂度为O(nlog(n) + m(log(n))2)——使用归并排序的情况下（预处理时从下往上处理就可以使用归并了）。

具体见代码

#include<bits/stdc++.h>using namespace std;typedef long long ll;const int maxn=1e6+50;const int maxm=1005;const ll mod=1000000007;vector<int> ln[maxn];vector<ll> sm[maxn];int len[maxn],n,q,a[maxn];void mergee(int x,int y){    int ad=len[y];    int lx=ln[x].size(),ly=ln[y].size();    int i=0,j=0,k=0;    while(i<lx&&j<ly){        if(ln[x][i]<ln[y][j]+ad)a[k++]=ln[x][i++];        else a[k++]=ln[y][j++]+ad;    }    while(i<lx)a[k++]=ln[x][i++];    while(j<ly)a[k++]=ln[y][j++]+ad;    for(int i=0;i<lx;i++)ln[x][i]=a[i];    for(int i=lx;i<k;i++)ln[x].push_back(a[i]);}ll cal(int node,int lx){    int lb=0,rb=ln[node].size()-1,mid;    while(lb<=rb){        mid=(lb+rb)>>1;        if(ln[node][mid]<=lx)lb=mid+1;        else rb=mid-1;    }    if(rb<0)return 0;    return 1ll*lx*(rb+1)-sm[node][rb];}int main(){    scanf("%d%d",&n,&q);    for(int i=2;i<=n;i++)scanf("%d",&len[i]);    for(int i=1;i<=n;i++)ln[i].push_back(0);//zero represent himself    for(int i=n;i>1;i--)mergee(i/2,i);    for(int i=1;i<=n;i++){        for(int j=0,k=ln[i].size();j<k;j++){            sm[i].push_back(ln[i][j]);            if(j)sm[i][j]+=sm[i][j-1];//prefix sum        }    }    for(int i=1;i<=q;i++){        int j,k;        scanf("%d%d",&j,&k);        ll cnt=0;        for(int last=0;j;k-=len[j],last=j,j>>=1){            if(k<0)break;            cnt+=k;            int lch=j<<1,rch=j<<1|1;            if(lch<=n&&last!=lch)                cnt+=cal(lch,k-len[lch]);            if(rch<=n&&last!=rch)                cnt+=cal(rch,k-len[rch]);        }        printf("%lld\n",cnt);    }    return 0;}