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.
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).
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 - an | = ans, 则有两种情况 p - an = ans , an - p = ans, 那么可以确定 an 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+7; const int INF = 1e6+7; struct node { int l, r, s; }tree[maxn * 55]; 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) >> 1; if(pos <= mid) { update(l, mid, tree[pre].l, tree[cur].l, pos); } else { update(mid + 1, 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) >> 1; int res = 0; if(x <= mid) { res += query(x, y, l ,mid, tree[pre].l, tree[cur].l); } if(y > mid) { res += query(x, y, mid + 1, r, tree[pre].r, tree[cur].r); } return res; } int main() { int T; scanf("%d", &T); while(T--) { cnt = 0; scanf("%d %d", &n, &m); for(int i = 1; i <= n; i++) { scanf("%d", &arr[i]); update(1, INF, root[i - 1], root[i], arr[i]); } int ans = 0; 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 = 0, pr = INF; while(pl <= pr) { int mid = (pl + pr) >> 1; if(query(max(1, p - mid), min(INF, p + mid), 1, INF, root[l - 1], 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 - 1; } else { pl = mid + 1; } } printf("%d\n", ans); } } return 0; }