主席树 - 查询某区间第 K 大

最新推荐文章于 2022-01-26 19:46:02 发布

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最新推荐文章于 2022-01-26 19:46:02 发布

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原文链接：http://www.cnblogs.com/ccut-ry/p/8831226.html

解决区间第K大问题，利用主席树数据结构处理大量查询请求。对于给定数组，快速找出指定区间的第K个最小元素。

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 10 ⁹ 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

5
6
3

Hint

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

题意：给你一维的一串数，询问区间第 K 大，主席树板子题

#define ll long long
const int maxn = 1e5+5;
const int mod = 1e9+7;
const double eps = 1e-9;
const double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;

int n, m;
int pre[maxn];
int rank[maxn];
int cnt, ss;
int root[maxn];
struct node
{
    int l, r;
    int sum;
}t[maxn*20];

void init(){
    cnt = 1;
    root[0] = 0;
    t[0].l = t[0].r = t[0].sum = 0;        
}

void update(int num, int &rt, int l, int r){
    t[cnt++] = t[rt];
    rt = cnt-1;
    t[rt].sum++;
    if (l == r) return;
    int m = (l+r)>>1;
    if (num <= m) update(num, t[rt].l, l, m);
    else update(num, t[rt].r, m+1, r);
}

int query(int i, int j, int k, int l, int r){
    int d = t[t[j].l].sum - t[t[i].l].sum;
    int m = (l+r)>>1;
    
    if (l == r) return l;
    if (k <= d) return query(t[i].l, t[j].l, k, l, m);
    else return query(t[i].r, t[j].r, k-d, m+1, r);
}

int main() {
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    cin >> n >> m;
    for(int i = 1; i <= n; i++){
        scanf("%d", &pre[i]);
        rank[i] = pre[i];
    }
    sort(rank+1, rank+1+n);
    ss = unique(rank+1, rank+1+n)-rank;
    //printf("--  %d\n", cnt);
    init();
    for(int i = 1; i <= n; i++){
        int x = lower_bound(rank+1, rank+ss, pre[i])-rank;
        root[i] = root[i-1];
        update(x, root[i], 1, n);
    }
    int a, b, c;
    for(int i = 1; i <= m; i++){
        scanf("%d%d%d", &a, &b, &c);
        int ans = query(root[a-1], root[b], c, 1, n);
        printf("%d\n", rank[ans]);
    }
    return 0;
}

转载于:https://www.cnblogs.com/ccut-ry/p/8831226.html