[Codeforces 894D. Ralph And His Tour in Binary Country]预处理+二分

[Codeforces 894D. Ralph And His Tour in Binary Country]预处理+二分

分类：Tree binary search

1. 题目链接

[Codeforces 894D. Ralph And His Tour in Binary Country]

2. 题意描述

有一棵 n 个节点的二叉树。第 i 条边连接着点 i+1 和点 ⌊i+12⌋ ，边长为 Li 。有 m 个询问，对于每个询问 j ，包含两个数 uj,hj 。对于所有到 uj 的距离小于 hj 的顶点 v ，求 ∑(hj−v)
数据范围： (1 ≤ n ≤ 106,1 ≤ m ≤ 105).

3. 解题思路

因为树比较特殊，是一棵完全二叉树。每个节点的深度最多是 logn 。那么可以预处理出每棵子树到该子树根节点的距离。排序之后，维护前缀和。
那么查询节点 uj 的时候，枚举 uj 到根节点的路径上的所有顶点。即可统计出答案。

4. 实现代码

#include 

using namespace std;

typedef long long ll;
typedef long double lb;
typedef unsigned int uint;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef pair pll;
typedef pair puu;
typedef pair pbb;
typedef vector<int> vi;

const int inf = 0x3f3f3f3f;
const ll infl = 0x3f3f3f3f3f3f3f3fLL;

#ifdef ___LOCAL_WONZY___
void debug() { cout << endl; }
template<typename T, typename ...R> void debug(T f, R ...r) {
    cout << "[" << f << "]";
    debug(r...);
}
#endif // ___LOCAL_WONZY___

const int MAXN = 1000006;
int n, m, u; ll h;
ll L[MAXN];
vector dep[MAXN], pre[MAXN];

void dfs(int rt) {
    if (rt > n) return;
    dep[rt].clear(); pre[rt].clear();
    dep[rt].push_back(0);
    int lch = rt << 1, rch = rt << 1 | 1;
    dfs(lch); dfs(rch);

    int i = 0, j = 0;
    if (lch <= n && rch <= n) {
        while (i < dep[lch].size() && j < dep[rch].size()) {
            ll a = dep[lch][i] + L[lch];
            ll b = dep[rch][j] + L[rch];
            if (a <= b) dep[rt].push_back(a), ++i;
            else dep[rt].push_back(b), ++j;
        }
    }
    while (lch <= n && i < dep[lch].size()) dep[rt].push_back(dep[lch][i++] + L[lch]);
    while (rch <= n && j < dep[rch].size()) dep[rt].push_back(dep[rch][j++] + L[rch]);
    // for(auto& e : dep[lch]) dep[rt].push_back(e + L[lch]);
    // for(auto& e : dep[rch]) dep[rt].push_back(e + L[rch]);
    // sort(dep[rt].begin(), dep[rt].end());
    pre[rt].resize(dep[rt].size());
    pre[rt][0] = dep[rt][0];
    for (int i = 1; i < dep[rt].size(); ++i) {
        pre[rt][i] = pre[rt][i - 1] + dep[rt][i];
    }
}

int main() {
#ifdef ___LOCAL_WONZY___
    freopen("input.txt", "r", stdin);
#endif // ___LOCAL_WONZY___
    scanf("%d %d", &n, &m);
    for (int i = 2; i <= n; ++ i) scanf("%lld", &L[i]);
    dfs(1);
    for (int i = 1; i <= m; ++ i) {
        scanf("%d %lld", &u, &h);
        int pos = lower_bound(dep[u].begin(), dep[u].end(), h) - dep[u].begin();
        ll ans = 0;
        if (pos != 0) ans += pos * h - pre[u][pos - 1];
        while (u > 1 && h - L[u] > 0) {
            h -= L[u]; ans += h;
            int v = u ^ 1;
            if (h - L[v] > 0) {
                ll hx = h - L[v];
                pos = lower_bound(dep[v].begin(), dep[v].end(), hx) - dep[v].begin();
                if (pos != 0) ans += pos * hx - pre[v][pos - 1];
            }
            u >>= 1;
        }
        printf("%lld\n", ans);
    }
#ifdef ___LOCAL_WONZY___
    cout << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC * 1000 << "ms." << endl;
#endif // ___LOCAL_WONZY___
    return 0;
}