POJ Lost Cows

【题解】

       参考https://blog.csdn.net/acmer_hades/article/details/46272605。设置数组pre_smaller，其中第i个元素即为输入的第i项，则显然pre_smaller[1] = 0。build_tree建立树结构，分解区间[1, N]，其中每个节的度要么为2，要么为0。query中参数smaller_num表示自身以及队列前方比自己编号小的cows的总数。确实是非常简洁的方法，将线段数组运用得淋漓尽致！不愧是NOI大神

【代码】

#include <iostream>
#include <cstdlib>
using namespace std;

#define maxn 100001

int N;
int pre_smaller[maxn], ans[maxn];

struct node 
{
    int left_val, right_val, len;
}s[4*maxn];

void build_tree(int root, int left_val, int right_val)
{
    s[root].left_val = left_val;
    s[root].right_val = right_val;
    s[root].len = right_val - left_val + 1;
    if (left_val == right_val)return;
    build_tree(2 * root, left_val, (left_val + right_val) / 2);
    build_tree(2 * root + 1, 1 + (left_val + right_val) / 2, right_val);
}

int query(int root, int smaller_num)
{
    s[root].len--;
    if (s[root].left_val == s[root].right_val)return s[root].left_val;
    else if (s[2 * root].len >= smaller_num)return query(2 * root, smaller_num);
    else return query(2 * root + 1, smaller_num - s[2 * root].len);
}

int main()
{
    cin >> N;
    for (int i = 2; i <= N; i++)cin >> pre_smaller[i];
    pre_smaller[1] = 0;
    build_tree(1, 1, N);
    for (int i = N; i >= 1; i--)ans[i] = query(1, pre_smaller[i] + 1);
    for (int i = 1; i <= N; i++)printf("%d
", ans[i]);
    return 0;
}