D. Ralph And His Tour in Binary Country
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http://codeforces.com/problemset/problem/894/D
给出一个平衡的二叉树,带边权,每组询问等价于求到A点距离不超过H的所有点到A点的距离和。
思考为什么是“平衡二叉树”,其具有层数不超过logn,只有每个节点最多有两个子的特点。
对每个点将其子树所有点到该点的距离维护成一个有序表,由于是静态的,排序即可。之后就可以使用lower_bound去找有多少个点满足条件,维护有序表的前缀和就可得解。
除了点A所在子树外,要沿着父节点向上爬,对路径上的每个节点及该节点的另一个子树求解,向上爬的过程中H是单调递减的。
思路是对的,结果维护有序表的时候没想清楚,用了动态开点的线段树,结果时间和空间都炸了。
下次出思路以后要想清楚需要维护的信息的本质,尝试用更简单的结构和方法去优化。
#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>#include <queue>#include <vector>using namespace std;typedef long long lint;#define maxn (1000010)#define limit (10000000)int L[maxn], n, m;vector<lint> a[maxn], sum[maxn];struct node { int v; lint d; node(int _v, lint _d) { v = _v; d = _d; }};queue<node> que;void build(int x){ while(!que.empty()) que.pop(); que.push(node(x, 0)); while(!que.empty()) { node now = que.front(); que.pop(); a[x].push_back(now.d); int y = now.v; if((y<<1) <= n && now.d + L[y<<1] <= limit) { que.push(node(y<<1, now.d + L[y<<1])); } if((y<<1|1) <= n && now.d + L[y<<1|1] <= limit) { que.push(node(y<<1|1, now.d + L[y<<1|1])); } } sort(a[x].begin(), a[x].end()); for(int i = 0, sz = a[x].size(); i < sz; i++) { sum[x].push_back(a[x][i]); if(i > 0) sum[x][i] += sum[x][i-1]; }}lint sol(int x, int k){ //printf("x = %d, k = %d\n", x, k); lint ans = 0; int p = lower_bound(a[x].begin(), a[x].end(), k) - a[x].begin(); if(p == a[x].size() || a[x][p] > k) p--; if(p >= 0) ans += (lint)k * (p + 1) - sum[x][p]; //printf("ans0 = %lld\n", ans); while(x >> 1) { k -= L[x]; if(k <= 0) break; ans += k; int y = x ^ 1; p = lower_bound(a[y].begin(), a[y].end(), k - L[y]) - a[y].begin(); if(p == a[y].size() || a[y][p] > k - L[y]) p--; if(p >= 0) ans += (lint)(k - L[y]) * (p + 1) - sum[y][p]; //printf("ans = %lld\n", ans); x >>= 1; } return ans;}int main(){ cin >> n >> m; for(int i = 2; i <= n; i++) { scanf("%d", &L[i]); } for(int i = 1; i <= n; i++) { build(i); } for(int i = 0, A, H; i < m; i++) { scanf("%d%d", &A, &H); printf("%I64d\n", sol(A, H)); } return 0;}
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