Codeforces Round #447 (Div. 2) D. Ralph And His Tour in Binary Country [DP+空间优化]_codeforces #round 447div2 d

本文介绍了一种在满二叉树上进行高效查询的方法，通过预处理每个子树内的距离信息，实现快速计算从指定起点到所有可达点的幸福值总和，特别适用于大规模数据集。

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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 $\text{[math]}$ (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
inputCopy
2 2
5
1 8
2 4
outputCopy
11
4
inputCopy
6 4
2
1
1
3
2
2 4
1 3
3 2
1 7
outputCopy
11
6
3
28

Note

Here is the explanation for the second sample.

Ralph's first query is to start tours from city 2 and Hi equals to 4. Here are the options:

He can choose city 5 as his terminal city. Since the distance between city 5 and city 2 is 3, he can gain happiness 4 - 3 = 1.
He can choose city 4 as his terminal city and gain happiness 3.
He can choose city 1 as his terminal city and gain happiness 2.
He can choose city 3 as his terminal city and gain happiness 1.
Note that Ralph can choose city 2 as his terminal city and gain happiness 4.
Ralph won't choose city 6 as his terminal city because the distance between city 6 and city 2 is 5, which leads to negative happiness for Ralph.

So the answer for the first query is 1 + 3 + 2 + 1 + 4 = 11.

题意:一颗满二叉树n(1e6)个顶点,边权都知道,一共q(1e5)次询问.从ST点出发有生命值H,求dis(st,任意一点x)<=H 的 sgima(H-dis)

思路:

1.对于一个点,出发有三种情况,左儿子/右儿子/父亲,求是否能走到以及走到后有多少个可能的值(二分)

2.那么从ST出发往左右儿子走的情况我们能都处理出来,但是往上走的我们还没处理. 再想,ST/2就到父亲来了,那岂不是h-cost[father]就是当前剩余的生命值了吗? 如果前面是左儿子过来的,只检查father的rson. 如果是右儿子爬过来的,那么只检查father的lson.

画图理解一下! 其实模拟一下,观察是可以做的?

时间复杂度:[n(1+1/2+1/4+...)=O(n) ]+ O(nlogn * log(logn))+O(q*logn*log(logn))

这道题还要注意空间复杂度: nlogn n=1e6. 正常开会爆内存,开vector优化空间

#include<bits/stdc++.h>
#define PI acos(-1.0)
using namespace std;
typedef long long ll;

const int N=1e6+6;
const int MOD=1e9+7;
const int INF=0x3f3f3f3f;

vector <int> ans[N];
vector <ll> sum[N];
int cost[N],h;
int n,q,st,last;

/// distance bewteen subtree root and subtree node
void dfs(int now,int rt,int fdis){
    if(now>n)   return ;
    int lson=now*2,rson=now*2+1;
    if(lson<=n) ans[rt].push_back(cost[lson]+fdis),dfs(lson,rt,cost[lson]+fdis);
    if(rson<=n) ans[rt].push_back(cost[rson]+fdis),dfs(rson,rt,cost[rson]+fdis);
}


int main(void){
    cin >> n>>q;
    for(int i=2;i<=n;i++)   scanf("%d",&cost[i]);
    for(int i=1;i<=n;i++) // pb(0)是为了符合平时写前缀和的习惯,把VEC第0个位置先占了,dfs是用来求当前节点到字节点的ALLDistans
        ans[i].push_back(0),dfs(i,i,0);

    for(int i=1;i<=n;i++){
        sort(ans[i].begin(),ans[i].end());
        sum[i].resize((int)ans[i].size()+1,0);
        for(int j=1;j<(int)ans[i].size();++j){
            sum[i][j]+=sum[i][j-1]+1ll*ans[i][j];
        }
    }

//    for(int i=1;i<=n;i++){
//        printf("%d: ",i);
//        for(auto j : ans[i])    printf("%lld ",j);
//        cout<<endl;
//        for(auto j : sum[i])    printf("%lld ",j.second);puts("*************");
//    }


    for(int i=1;i<=q;i++){
        scanf("%d%d",&st,&h);
        ll res=0;
        while(st>0 && h>0){
            res+=1ll*h;
            int lson=st*2;
            int rson=st*2+1;
            if(last!=lson && lson<=n){
                int te=h-cost[lson];
//                cout <<"te="<<te<<endl;
                if(te>0){
                    res+=te;
//                    cout <<"te1="<<te<<endl;
                    int pos=prev(lower_bound(ans[lson].begin(),ans[lson].end(),te))-ans[lson].begin();
//                    cout <<"pos="<<pos<<endl;
                    res+= (1ll*pos*te-sum[lson][pos]);
                }
//                cout <<"rse1="<<res<<endl;
            }
            if(last!=rson && rson<=n){
                int te=h-cost[rson];
                if(te>0){
                    res+=te;
                    int pos=prev(lower_bound(ans[rson].begin(),ans[rson].end(),te))-ans[rson].begin();
                    res+= (1ll*pos*te-sum[rson][pos]);
                }
//                cout <<"rse2="<<res<<endl;
            }
            last=st;
            h-=cost[last];
//            cout <<"h="<<h<<endl;
            st/=2;
//            cout <<"st="<<st<<endl;
        }
            cout << res  << endl;
    }


    return 0;
}
/*********
6 4
2
1
1
3
2
1 3

*********/