POJ 6621: K-th Closest Distance（主席树 + 二分）

K-th Closest Distance

Time Limit: 20000/15000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 1697    Accepted Submission(s): 633

Problem Description

You have an array: a1, a2, , an and you must answer for some queries.
For each query, you are given an interval [L, R] and two numbers p and K. Your goal is to find the Kth closest distance between p and aL, aL+1, ..., aR.
The distance between p and ai is equal to |p - ai|.
For example:
A = {31, 2, 5, 45, 4 } and L = 2, R = 5, p = 3, K = 2.
|p - a2| = 1, |p - a3| = 2, |p - a4| = 42, |p - a5| = 1.
Sorted distance is {1, 1, 2, 42}. Thus, the 2nd closest distance is 1.

 

Input

The first line of the input contains an integer T (1 <= T <= 3) denoting the number of test cases.
For each test case:
冘The first line contains two integers n and m (1 <= n, m <= 10^5) denoting the size of array and number of queries.
The second line contains n space-separated integers a1, a2, ..., an (1 <= ai <= 10^6). Each value of array is unique.
Each of the next m lines contains four integers L', R', p' and K'.
From these 4 numbers, you must get a real query L, R, p, K like this: 
L = L' xor X, R = R' xor X, p = p' xor X, K = K' xor X, where X is just previous answer and at the beginning, X = 0.
(1 <= L < R <= n, 1 <= p <= 10^6, 1 <= K <= 169, R - L + 1 >= K).

 

Output

For each query print a single line containing the Kth closest distance between p and aL, aL+1, ..., aR.

 

Sample Input

1
5 2
31 2 5 45 4
1 5 5 1
2 5 3 2

 

Sample Output

0
1

 

 

杭电多校第四场...

结构：主席树 + 二分

 

题解：首先按照模板走，创建一颗主席树。然后更具题目意思，我先设ans就是我二分的时候要找到的答案，那么就有 | p - a_n | = ans, 则有两种情况 p - a_n = ans , a_n - p = ans, 那么可以确定 a_n 的值域 [ p - ans, p + ans ]。

 

注意：主席树的数组一定要开的够大，不然就会TLE...我开的40倍就TLE了，看大佬的开的55倍，改了一下就过了。

 

#include <iostream>
#include <cstdio>

using namespace std;

const int maxn = 1e5+;
const int INF = 1e6+;

struct node {
    int l, r, s;
}tree[maxn * ];   

int root[maxn];
int arr[maxn];
int n, m;
int cnt;

void update(int l, int r, int pre, int &cur, int pos) {
    cur = ++cnt;
    tree[cur] = tree[pre];
    tree[cur].s++;
    if(l == r) {
        return;
    }
    int mid = (l + r) >> ;
    if(pos <= mid) {
        update(l, mid, tree[pre].l, tree[cur].l, pos);
    } else {
        update(mid + , r, tree[pre].r, tree[cur].r, pos);
    }
}

int query(int x, int y, int l ,int r, int pre, int cur) {
    if(x <= l && r <= y) {
        return tree[cur].s - tree[pre].s;
    }
    int mid = (l + r) >> ;
    int res = ;
    if(x <= mid) {
        res += query(x, y, l ,mid, tree[pre].l, tree[cur].l);
    }
    if(y > mid) {
        res += query(x, y, mid + , r, tree[pre].r, tree[cur].r);
    }
    return res;
}

int main() {
    int T;
    scanf("%d", &T);
    while(T--) {
        cnt = ;
        scanf("%d %d", &n, &m);
        for(int i = ; i <= n; i++) {
            scanf("%d", &arr[i]);
            update(, INF, root[i - ], root[i], arr[i]);
        }
        int ans = ;
        while(m--) {
            int l, r, p, k;
            scanf("%d %d %d %d", &l, &r, &p, &k);
            l = l ^ ans;
            r = r ^ ans;
            p = p ^ ans;
            k = k ^ ans;
            int pl = , pr = INF;
            while(pl <= pr) {
                int mid = (pl + pr) >> ;
                if(query(max(, p - mid), min(INF, p + mid), , INF, root[l - ], root[r]) >= k) {      //此处的max和min是为了防止超出[0, 1e6]的范围
                    ans = mid;
                    pr = mid - ;
                } else {
                    pl = mid + ;
                }
            }
            printf("%d\n", ans);
        }
    }
    return ;
}