Codeforces Round #447 (Div. 2) D 预处理+归并+ 二分

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

Before answering each query, pre-process on the tree. On each vertice, we can get a sorted array of all the vertices in its subtree sorted by distance to this vertex. And it costs O(nlog(n)) time using merge sort or O(n(log(n))2) time using std::sort. If you use std::sort, you should implement it carefully or it won’t be able to fit in the time limit. Because the tree is an almost complete binary tree, one vertex will appear at most [log(n)] times in all n sorted arrays,so the memory complexity is O(nlog(n)).

To answer each query, we can iterate on the highest vertex on the tour and do binary search on the sorted array to get the answer. We’ll do at most [log(n)] times of iteration and the binary search is O(log(n)) per iteration, so we can answer each query in O((log(n))2) time.

Overall, the time complexity is O(nlog(n) + m(log(n))2) and the memory complexity is O(nlog(n)). If you use std::sort, the time complexity will be O((n + m)(log(n))2) and the memory complexity is the same.

题意

有n个城市，i和2i存在一条边，i和2i+1也连一条边。
给你每条边的长度。
现在给你m个询问，每个询问给你x和h，表示起点为x，然后让你求sigma(max(h-dist[i],0))

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>

using namespace std;
typedef long long LL;
const int N=1000050;

vector<LL> sm[N];
vector<int> ln[N];
int len[N],n,q,a[N];
inline void merge(int x,int y){
    register int i=0,j=0,k=0,ad=len[y];
    register int lx=ln[x].size(),ly=ln[y].size();
    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(i=0;i<lx;i++)ln[x][i]=a[i];
    for(i=lx;i<k;i++)
        ln[x].push_back(a[i]);
}
LL cal(int nod,int lx){
    register int lb=0,rb=ln[nod].size()-1,mid;
    while(lb<=rb){
        mid=lb+rb>>1;
        if(ln[nod][mid]<=lx)
            lb=mid+1;
        else
            rb=mid-1;
    }
    if(rb<0)return 0;
    return 1ll*lx*(rb+1)-sm[nod][rb];
}

int main(){

    int i,j,k,las;
    scanf("%d%d",&n,&q);
    for(i=2;i<=n;i++)scanf("%d",&len[i]);
    for(i=1;i<=n;i++)
        ln[i].push_back(0);
    for(i=n;i>1;i--)merge(i/2,i);
    for(i=1;i<=n;i++){
        for(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];
        }
    }
    for(i=1;i<=q;i++){
        scanf("%d%d",&j,&k);
        register LL cnt=0;
        for(las=0;j;k-=len[j],las=j,j>>=1){
            if(k<0)break;cnt+=k;
            int lch=j<<1,rch=j<<1|1;
            if(lch<=n&&las!=lch)
                cnt+=cal(lch,k-len[lch]);
            if(rch<=n&&las!=rch)
                cnt+=cal(rch,k-len[rch]);
        }
        printf("%lld\n",cnt);
    }

    return 0;
}