poj 2104 （划分树模板）

Description

You are working for Macrohard company in data structures department. After failing your previous task about key insertion you were asked to write a new data structure that would be able to return quickly k-th order statistics in the array segment.
That is, given an array a[1...n] of different integer numbers, your program must answer a series of questions Q(i, j, k) in the form: "What would be the k-th number in a[i...j] segment, if this segment was sorted?"
For example, consider the array a = (1, 5, 2, 6, 3, 7, 4). Let the question be Q(2, 5, 3). The segment a[2...5] is (5, 2, 6, 3). If we sort this segment, we get (2, 3, 5, 6), the third number is 5, and therefore the answer to the question is 5.

Input

The first line of the input file contains n --- the size of the array, and m --- the number of questions to answer (1 <= n <= 100 000, 1 <= m <= 5 000).
The second line contains n different integer numbers not exceeding 10⁹ by their absolute values --- the array for which the answers should be given.
The following m lines contain question descriptions, each description consists of three numbers: i, j, and k (1 <= i <= j <= n, 1 <= k <= j - i + 1) and represents the question Q(i, j, k).

Output

For each question output the answer to it --- the k-th number in sorted a[i...j] segment.

Sample Input

7 3

1 5 2 6 3 7 4

2 5 3

4 4 1

1 7 3

Sample Output

Hint

This problem has huge input,so please use c-style input(scanf,printf),or you may got time limit exceed.

今天主要来学习一个划分树，也就是入门吧，我先拓宽一下知识面吧，

感觉只有拓宽算法知识面才能更加深入学习更复杂的算法

划分树板子

 #include <cstdio>

 #include <algorithm>

 #include <cstring>

 #include <cmath>

 using namespace std;

 typedef long long LL;

 const int maxn = 1e5 + ;

 int a[maxn], sorted[maxn];

 int num[][maxn], val[][maxn];

 void build(int l, int r, int ceng) {

     if (l == r) return ;

     int mid = (l + r) >> , same = mid - l + ;

     for (int i = l ; i <= r ; i++)

         if (val[ceng][i] < sorted[mid]) same--;

     int ln = l, rn = mid + ;

     for (int i = l ; i <= r ; i++) {

         if (i == l) num[ceng][i] = ;

         else num[ceng][i] = num[ceng][i - ];

         if (val[ceng][i] < sorted[mid] || val[ceng][i] == sorted[mid] && same > ) {

             val[ceng + ][ln++] = val[ceng][i];

             num[ceng][i]++;

             if (val[ceng][i] == sorted[mid]) same--;

         } else val[ceng + ][rn++] = val[ceng][i];

     }

     build(l, mid, ceng + );

     build(mid + , r, ceng + );

 }

 int query(int ceng, int L, int R, int l, int r, int k) {

     if (L == R)  return val[ceng][L];

     int lsum;

     if (l == L) lsum = ;

     else lsum = num[ceng][l - ];

     int tot = num[ceng][r] - lsum;

     if (tot >= k) return query(ceng + , L, (L + R) / , L + lsum, L + num[ceng][r] - , k);

     else {

         int lr = (L + R) /  +  + (l - L - lsum);

         return query(ceng + , (L + R) /  + , R, lr, lr + r - l +  - tot - , k - tot);

     }

 }

 int main() {

     int n, m, l, r, k;

     while(scanf("%d%d", &n, &m) != EOF) {

         for (int i =  ; i <= n ; i++) {

             scanf("%d", &val[][i]);

             sorted[i] = val[][i];

         }

         sort(sorted + , sorted + n + );

         build(, n, );

         while(m--) {

             scanf("%d%d%d", &l, &r, &k);

             printf("%d\n", query(, , n, l, r, k ));

         }

     }

     return ;

 }

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poj 2104 （划分树模板）

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