标签：fill exists rtu sum option index rom binary The
http://codeforces.com/contest/985/problem/E
Mishka received a gift of multicolored pencils for his birthday! Unfortunately he lives in a monochrome world, where everything is of the same color and only saturation differs. This pack can be represented as a sequence a1,?a2,?...,?an of n integer numbers — saturation of the color of each pencil. Now Mishka wants to put all the mess in the pack in order. He has an infinite number of empty boxes to do this. He would like to fill some boxes in such a way that:
Help Mishka to determine if it‘s possible to distribute all the pencils into boxes. Print "YES" if there exists such a distribution. Otherwise print "NO".
The first line contains three integer numbers n, k and d (1?≤?k?≤?n?≤?5·105, 0?≤?d?≤?109) — the number of pencils, minimal size of any non-empty box and maximal difference in saturation between any pair of pencils in the same box, respectively.
The second line contains n integer numbers a1,?a2,?...,?an (1?≤?ai?≤?109) — saturation of color of each pencil.
6 3 10
7 2 7 7 4 2YES有n个数字，将他们分成若干组，每组至少k个元素，组内元素差不超过d.
There are n numbers, devide them into many groups, on which there are at least k elements.
If x,y in the same groups, they fits \(|x-y|<d\).
首先将元素按大小排序。dp[i]表示前i个可以满足条件。转移方程: dp[i] = dp[j]=1?1:0; 其中\(j<i-k\)，\(a[i]-a[j]<=d\). 我们用树状数组维护区间，用二分查找去找最小的j。
sort the elements for the value. dp[i] represent former i elements can fit the condition. The transform equation is: dp[i] = dp[j]=1?1:0; (\(j<i-k\)，\(a[i]-a[j]<=d\)). We use BIT to maintain the segment, and use binary search to find the minimum j.
#include <bits/stdc++.h>
using namespace std;
int n, k, d;
int a[500010];
int dp[500010];
int c[500010];
void update(int k) {
while (k <= n) {
c[k]++;
k |= k + 1;
}
};
int sum(int k) {
int ret = 0;
while (k >= 0) {
ret += c[k];
k = ((k + 1) & k) - 1;
}
return ret;
}
int get(int l, int r) {
if (l > r) return 0;
return sum(r) - sum(l - 1);
}
int main() {
cin >> n >> k >> d;
for (int i = 1; i <= n; i++)
cin >> a[i];
sort(a + 1, a + n + 1, less<int>());
int l = 0;
dp[0] = 1;
update(0);
for (int i = 1; i <= n; i++) {
while (l < i && a[i] - a[l] > d) l++;
dp[i] = get(l - 1, i - k) >= 1 ? 1 : 0;
if (dp[i]) update(i);
}
cout << (dp[n] ? "YES" : "NO") << endl;
}Codeforces Educational Codeforces Round 44 (Rated for Div. 2) E. Pencils and Boxes
标签：fill exists rtu sum option index rom binary The
原文地址：https://www.cnblogs.com/EDGsheryl/p/9165357.html
