【树状数组】CF961E Tufurama

挺巧妙的数据结构题（不过据说这是一种套路？

E. Tufurama

One day Polycarp decided to rewatch his absolute favourite episode of well-known TV series "Tufurama". He was pretty surprised when he got results only for season 7 episode 3 with his search query of "Watch Tufurama season 3 episode 7 online full hd free". This got Polycarp confused — what if he decides to rewatch the entire series someday and won't be able to find the right episodes to watch? Polycarp now wants to count the number of times he will be forced to search for an episode using some different method.

TV series have n seasons (numbered 1 through n), the i-th season has ai episodes (numbered 1 through ai). Polycarp thinks that if for some pair of integers x and y (x < y) exist both season x episode y and season y episode x then one of these search queries will include the wrong results. Help Polycarp to calculate the number of such pairs!

Input

The first line contains one integer n (1 ≤ n ≤ 2·10^5) — the number of seasons.

The second line contains n integers separated by space a1, a2, ..., an (1 ≤ ai ≤ 10^9) — number of episodes in each season.

Output

Print one integer — the number of pairs x and y (x < y) such that there exist both season x episode y and season y episode x.

题目大意

有一部电视剧有n季，每一季有ai集。定义二元组(i,j)：存在第i季有第j集。求(i,j)与(j,i)同时合法(i<j)的对数。

真实题意就是：求<i,j>对数，使得a[i]≥j，a[j]≥i并且(i<j)

看上去很可做的样子，对吧……

题目分析

基础的暴力

从1..n季，每一季都分别判断对答案的贡献。

例如对于 4 \n 3 5 1 2 ，依次检查(1,2)是否存在(2,1)；(1,3)是否存在(3,1)……

首先发现a[i]对于答案的贡献最大也就是到n为止，那么读入时候先取个min(n)。

考虑一下check()是O(n)的，所以总复杂度是O(n²)的。

BIT做法

像很多其他题一样，对于这样的、关于元素大小关系之间的限制的题目，先排个序总是能够解决个一维限制掉去的。

我们使用一个结构体node x，x.i表示季数；x.a表示该季的集数。首先对x.a排序。那么就变成这个样子：

p[].a(j)　　3　　5　　1　　2
p[].i(i)　　1　　2　　3　　4
　　|
　　|
p[].a(j)　　1　　2　　3　　4 (取min之后)
p[].i(i)　　3　　4　　1　　2

先考虑每次的统计，那么只要ans+=query(a[i])就可以了。意思就是说ans加上1..a[i]季的贡献（其中每一季的贡献要么是0要么是1，但是由于之后会有修改，所以我们用BIT维护）

接着考虑修改，设立一个now指向当前最小合法的p[]。这个now用来更新那些已经过气没有贡献的答案。

这里「没有贡献的答案」指的是p[now].a<i的情况。说人话就是p[now]的电视剧集数太小了，已经不会再有贡献了，因此now++，判断下一个p[]是否可能会对答案有贡献。个人感觉有那么一点相似单调队列和 wqs二分的情况（但是我不是非常清楚）？

为了去除这些没有贡献的季数的影响，我们只需将p[now].i位置在树状数组上-1即可。意思是说这个季数在之后的统计上都不会有贡献了。

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 long long ans;
 4 int n,now,a[200035];
 5 struct node
 6 {
 7     int a,i;
 8     bool operator < (node &xx) const
 9     {
10         return a < xx.a;
11     }
12 }p[200035];
13 int f[200035];
14 int read()
15 {
16     char ch = getchar();int num = 0;
17     for (; !isdigit(ch); ch = getchar());
18     for (; isdigit(ch); ch = getchar())
19         num = (num<<3)+(num<<1)+ch-48;
20     return num;
21 }
22 int lowbit(int x){return x&-x;}
23 void add(int x, int c){for (; x<=n+1; x+=lowbit(x))f[x]+=c;}
24 int query(int x)
25 {
26     int ret = 0;
27     for (; x; x-=lowbit(x))
28         ret += f[x];
29     return ret;
30 }
31 int main()
32 {
33     n = read();now = 1;
34     for (int i=1; i<=n; i++)
35         a[i] = min(read(), n+1), p[i].a = a[i], p[i].i = i, add(i, 1);
36     sort(p+1, p+n+1);
37     for (int i=1; i<=n; i++)
38     {
39         while (now<=n && p[now].a < i)add(p[now++].i, -1);
40         ans += query(a[i]);
41         if (a[i] >= i)ans--;
42     }
43     cout << ans / 2 << endl;
44     return 0;
45 }