Codeforces 894D(DFS+二分)

Codeforces 894D
题意：给出一颗带边权完全二叉树，有$m$个询问，每个询问有$A_i,H_i$，求从$A_i$出发分别到其他点$x$的$H_i-dis(A_i,x)$和

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#define ms(i, j) memset(i, j, sizeof i)
#define LL long long
using namespace std;
LL n, m, L[1000000 + 5]; 
vector<LL> dis[1000000 + 5], qzh[1000000 + 5];
LL query(LL a, LL h) {
    if (h < 0) return 0;
    LL p = upper_bound(dis[a].begin(), dis[a].end(), h) - dis[a].begin();
    return p * h - qzh[a][p - 1];
}
void clean() {}
void solve() {
    clean();
    for (LL i = 2; i <= n; i++) scanf("%I64d", &L[i]);
    for (LL x = n; x >= 1; x--) {
        dis[x].push_back(0);
        LL lc = (x << 1), rc = (x << 1 | 1);
        if (lc <= n) for (LL i = 0; i < (LL)dis[lc].size(); i++) dis[x].push_back(dis[lc][i] + L[lc]);
        if (rc <= n) for (LL i = 0; i < (LL)dis[rc].size(); i++) dis[x].push_back(dis[rc][i] + L[rc]);
        sort(dis[x].begin(), dis[x].end());
        qzh[x].resize(dis[x].size());
        qzh[x][0] = 0;
        for (LL i = 1; i < qzh[x].size(); i++) qzh[x][i] = qzh[x][i - 1] + dis[x][i];
    }
    while (m--) {
        LL a, h; scanf("%I64d%I64d", &a, &h);
        LL ans = 0, lst = 0;
        while (a) {
            if (h < 0) break;
            ans += h;
            LL lc = (a << 1), rc = (a << 1 | 1);
            if (lc <= n && lst != lc) ans += query(lc, h - L[lc]);
            if (rc <= n && lst != rc) ans += query(rc, h - L[rc]);
            lst = a, h -= L[a], a >>= 1;
        }
        printf("%I64d\n", ans);
    }
}
int main() {
    scanf("%I64d%I64d", &n, &m), solve();
    return 0;
}