Codeforces Round #312 div 2 C的二叉树实现

        感谢波兰的红名大牛Errichto的讲解。

        这道题中涉及到了两个操作，一个是加倍，另外一个除2并向下取整。这如果放在一个二叉树上的操作无非就是向上到达其父节点，向下（只能走左孩子）到达其子节点。我直接贴大牛的原话吧

Let's say input is n = 3 and array 2,5,6. Let's think about a binary tree with 7 (in my code there are always 2^17-1) nodes where node number x has children 2*x and 2*x+1. Root has number 1. Now let's think about it as a graph. There is one "thing" in node number 2, one in node number 5, one in node number 6. What is a node minimizing sum of distances to these "things"?

Note: we can only go up (to parent) or to the left son. Going up means dividing by 2 and going to left son means multiplying by 2.

现在的关键是找到这个节点，使得所有的元素到达它所需要的花费最小。

这里我们需要用上两个数组，一个数组res需要记录的是点x的子树到达其顶点x所需要的花费；一个数组count记录的是点x有多少个数能够达到。就拿上面的例子来说吧

count[5]=count[6]=1是因为这两个数本身就已经存在了，而count[2]=2的原因是count[5]是做为该节点的右孩子，这说明是可以到达的，后面count[1]=3,count[3]=1原因也是如此。

而另外一个数组res的值得求法应该是

res[x]=res[2*x]+res[2*x+1]+count[2*x]+count[2*x+1]

原因和其定义有关因为res[x]代表其子树到达其顶点的花费，而这些花费不能代表该顶点到达其父节点的花费，简而言之就是要加上自己的花费和它的子树到达它的花费。

#include<iostream>#include<cstring>#include<algorithm>#include<fstream>#include<cstring>#include<cstdio>using namespace std;typedef long long LL;#define mmax 128*1024LL res[mmax];LL cont[mmax];LL ans = (LL)mmax*mmax;void rec(int i, int RES, int COU){ans = min(ans, RES + res[i]);if (i > mmax / 2) return;int itsValue = cont[i] - cont[2 * i] - cont[2 * i + 1];rec(2 * i, RES + res[2 * i + 1] + 2 * cont[2 * i + 1] + itsValue + COU, COU + cont[2 * i + 1] + itsValue);if (res[2 * i] + itsValue + COU == 0)//当左支已经无法达到，这时跳转右支，算右支的子树rec(2 * i + 1, RES + res[2 * i] + 2 * cont[2 * i] + itsValue + COU, COU + cont[2 * i] + itsValue);}int main(){int n;scanf("%d", &n);for (int i = 0; i < n; i++){int a;scanf("%d", &a);cont[a]++;}for (int i = mmax / 2; i>0; i--){res[i] = res[i * 2] + res[i * 2 + 1] + cont[2 * i] + cont[2 * i + 1];cont[i] += cont[2 * i] + cont[2 * i + 1];}rec(1, 0, 0);cout << ans << endl;return 0;}