Codeforces Round #447 (Div. 2) D. Ralph And His Tour in Binary Country

题解 ：这个题目是一颗完全二叉树，根节点为1号节点， 在进行处理之前，我们可以对每个节点的子树节点以距离为关键字排序，思路类似于归并排序 （从底部向上面一层一层的排序） 当然排序之前我们需要把距离 0 加入这个排序的数组 （因为自身也可以加入进去，然后我们求一个前缀和，为了后面计算的方便） 然后进行计算 ，怎么算呢，每次以当前节点为根节点的的子树 （只算子孙节点对答案的贡献）就是二分控制距离就可以了，然后向自己的父亲节点走，再计算父亲节点为根对答案的贡献依次知道到控制距离小于等于0就可以了 输出答案就好了

#include <iostream>#include <algorithm>#include <cstring>#include <cstdio>#include <set>#include <vector>#define ll long longusing namespace std;const int maxn = 1e6 + 10;ll len[maxn] = {0};vector <ll> v[maxn];vector <ll> sum[maxn];ll temp[maxn] = {0};void merge (int node,int son) {    int sz1 = v[node].size(),sz2 = v[son].size();    ll length = len[son];    int i = 0,j = 0,k = 0;    while (i < sz1 && j < sz2) {        if (v[node][i] < v[son][j] + length) {            temp[k ++] = v[node][i ++];        }        else {            temp[k ++] = v[son][j ++] + length;        }    }    while (i < sz1) temp[k ++] = v[node][i ++];    while (j < sz2) temp[k ++] = v[son][j ++] + length;    for (i = 0;i < sz1; ++ i) v[node][i] = temp[i];    for (i = sz1;i < k; ++ i) v[node].push_back(temp[i]);}ll cal (int node,ll val) {    if (val <= 0) return 0;    ll pos = -1;    int mid;    int l = 0,r = v[node].size() - 1;    while (r >= l) {        mid = (l + r) >> 1;        if (v[node][mid] <= val) {            l = mid + 1;            pos = mid;        }        else {            r = mid - 1;        }    }    if (pos < 0) return 0;    return val * (pos + 1ll) - sum[node][pos];}int main () {    ios_base :: sync_with_stdio(false);    int n,Q;    cin >> n >> Q;    for (int i = 2;i <= n; ++ i) cin >> len[i];    for (int i = 1;i <= n; ++ i) v[i].push_back(0ll);    for (int i = n;i > 1; -- i) merge (i / 2,i);    for (int i = 1;i <= n; ++ i) {        int k = v[i].size();        sum[i].push_back(v[i][0]);        for (int j = 1;j < k; ++ j) {            sum[i].push_back(sum[i][j - 1] + v[i][j]);        }    }    while (Q --) {        int id,lim;        cin >> id >> lim;        ll ans = 0;        for (int las = 0;id;las = id,id = id / 2) {            if (lim <= 0) break;            ans += lim;            int l = id << 1,r = (id << 1) | 1;            if (l != las && l <= n) {                ans += cal (l,lim - len[l]);            }            if (r != las && r <= n) {                ans += cal (r,lim - len[r]);            }            lim -= len[id];        }        cout << ans << endl;    }    return 0;}