2017多校第二场 HDU 6047 Maximum Sequence 线段树或者multiset维护区间最值
来源:互联网 发布:nemo软件好用吗 编辑:程序博客网 时间:2024/06/08 05:58
Problem Description
Steph is extremely obsessed with “sequence problems” that are usually seen on magazines: Given the sequence 11, 23, 30, 35, what is the next number? Steph always finds them too easy for such a genius like himself until one day Klay comes up with a problem and ask him about it.
Given two integer sequences {ai} and {bi} with the same length n, you are to find the next n numbers of {ai}: an+1…a2n. Just like always, there are some restrictions on an+1…a2n: for each number ai, you must choose a number bk from {bi}, and it must satisfy ai≤max{aj-j│bk≤j}, and any bk can’t be chosen more than once. Apparently, there are a great many possibilities, so you are required to find max{∑2nn+1ai} modulo 109+7 .
Now Steph finds it too hard to solve the problem, please help him.
Input
The input contains no more than 20 test cases.
For each test case, the first line consists of one integer n. The next line consists of n integers representing {ai}. And the third line consists of n integers representing {bi}.
1≤n≤250000, n≤a_i≤1500000, 1≤b_i≤n.
Output
For each test case, print the answer on one line: max{∑2nn+1ai} modulo 109+7。
Sample Input
4
8 11 8 5
3 1 4 2
Sample Output
27
Hint
For the first sample:
1. Choose 2 from {bi}, then a_2…a_4 are available for a_5, and you can let a_5=a_2-2=9;
2. Choose 1 from {bi}, then a_1…a_5 are available for a_6, and you can let a_6=a_2-2=9;
解法:一个显然的贪心就是我们区间越长得到的最大值越大,所以我们先把b排序,从小到大找当前取的b[j]到当前i位置的a[i]-i的最大值,这显然可以用mulitiset或者线段树来维护,维护了b[j]之后要把b[j]到b[j+1]之前的值都删除掉,因为有重复元素所以用了multiset。
复杂度:O(nlogn)
#include <bits/stdc++.h>typedef long long LL;using namespace std;const LL huyu = 1e9 + 7;const LL maxn = 300000;LL a[maxn * 2];LL b[maxn];int main(){ LL n; while(~scanf("%lld", &n)) { for(LL i = 1; i <= n; i++) scanf("%lld", &a[i]); for(LL i = 1; i <= n; i++) scanf("%lld", &b[i]); sort(b + 1, b + 1 + n); multiset<LL> s; for(LL i = 1; i <= n; i++) s.insert(a[i] - i); LL pos = 1; LL ans = 0; for(LL i = 1; i <= n; i++) { for(; pos < b[i]; pos++) { s.erase(s.find(a[pos] - pos)); } LL tt = *s.rbegin(); ans += tt; ans %= huyu; s.insert(tt - n - i); } ans %= huyu; if(ans < 0) ans += huyu; printf("%lld\n", ans); } return 0;}
- 2017多校第二场 HDU 6047 Maximum Sequence 线段树或者multiset维护区间最值
- 2017 多校训练第二场 HDU 6047 Maximum Sequence
- CF 6E 线段树 or Multiset or 双端队列维护区间最值
- HDU 6047 Maximum Sequence 贪心 区间最值
- 2017 多校训练第二场 HDU 6047 Maximum Sequence(贪心+优先队列)
- 2017多校联合第二场 1003题 hdu 6047 Maximum Sequence O(n) (有理有据地)贪心
- [线段树 区间最值操作] HDU 5306 Gorgeous Sequence
- HDU 6047 Maximum Sequence (贪心,线段树)
- HDU 6047 Maximum Sequence(线段树)
- 2017 杭电多校联赛第二场 1003 Maximum Sequence(单调队列)HDU 6047
- HDU 1754 I Hate It(线段树维护 区间最值)
- HDU 6070 Dirt Ratio 分数规划 二分 线段树维护区间最值
- poj2823Sliding Window【线段树维护滚动区间最值】
- 51nod-1376(线段树维护区间最值)
- hdu 4902 Nice boat 多校第四场 线段树的区间置数+区间更新
- HDU6047 Maximum Sequence(贪心,2017 HDU多校联赛 第2场)
- HDU 4638 多校第四场1007 离线询问,树状数组||线段树维护
- 2017杭电多校联赛第二场-Maximum Sequence(hdu6047)
- HDOJ 6055-Regular polygon
- String,StringBuffer,StringBuilder 三者的讨论
- aop源码解析一:注册BPP
- ios底部输入框输入时被隐藏的bug解决
- 自己总结的html css知识点
- 2017多校第二场 HDU 6047 Maximum Sequence 线段树或者multiset维护区间最值
- 字符串替换
- Spanned及CharSequence
- c# Console.WriteLine() Console.ReadLine()
- aop源码解析二:postProcessBeforeInstantiation
- hibernate的get,load,persist方法比较(二)
- MySQL远程连接提示Accesss denied for user 'root'@'此处为你自己的ip'(using password:YES)
- 数据结构-排序
- makefile入门