洛谷P3031 [USACO11NOV]（中位数，树状数组）

题目地址：https://www.luogu.org/problemnew/show/3031

题目描述
Farmer John has lined up his N (1 <= N <= 100,000) cows in a row to measure their heights; cow i has height H_i (1 <= H_i <= 1,000,000,000) nanometers--FJ believes in precise measurements! He wants to take a picture of some contiguous subsequence of the cows to submit to a bovine photography contest at the county fair.
The fair has a very strange rule about all submitted photos: a photograph is only valid to submit if it depicts a group of cows whose median height is at least a certain threshold X (1 <= X <= 1,000,000,000).
For purposes of this problem, we define the median of an array A[0...K] to be A[ceiling(K/2)] after A is sorted, where ceiling(K/2) gives K/2 rounded up to the nearest integer (or K/2 itself, it K/2 is an integer to begin with). For example the median of {7, 3, 2, 6} is 6, and the median of {5, 4, 8} is 5.
Please help FJ count the number of different contiguous subsequences of his cows that he could potentially submit to the photography contest.
给出一串数字，问中位数大于等于X的连续子串有几个。（这里如果有偶数个数，定义为偏大的那一个而非中间取平均）
输入格式：
Line 1: Two space-separated integers: N and X.
Lines 2..N+1: Line i+1 contains the single integer H_i.
输出格式：
Line 1: The number of subsequences of FJ's cows that have median at least X. Note this may not fit into a 32-bit integer.
输入输出样例
输入样例#1： 复制
4 6 
10 
5 
6 
2 
输出样例#1： 复制
7 
说明
FJ's four cows have heights 10, 5, 6, 2. We want to know how many contiguous subsequences have median at least 6.
There are 10 possible contiguous subsequences to consider. Of these, only 7 have median at least 6. They are {10}, {6}, {10, 5}, {5, 6}, {6, 2}, {10, 5, 6}, {10, 5, 6, 2}.

【分析】：

dp的思想，树状数组优化。设数组dp[i]表示以第i元素结尾时，>=X的数量。

那么以第i元素结尾的符合条件的子序列要满足 大于X的数不少于小于X的数

则开头元素j需满足：2*(dp[i] - dp[j-1]) >= i - (j-1)

即  2*dp[i] - i >= 2*dp[j-1] - (j-1)

令 w = 2*dp[i] - i

只需结尾元素的 w 大于开头元素的 w ，树状数组维护即可

【代码】：

[cpp] view plain copy

1. #include<bits/stdc++.h>  
2. using namespace std;  
3. typedef long long ll;  
4. int k,n,x;  
5. int c[1010101];  
6. void add(int k,int num) //a[k]的值增加num  
7. {  
8.     while(k<=2*n+1)  
9.     {  
10.         c[k]+=num;  
11.         k+=k&-k;  
12.     }  
13. }  
14. ll read(int k)//1~k的区间和  
15. {  
16.     ll sum=0;  
17.     while(k)  
18.     {  
19.         sum+=c[k];  
20.         k-=k&-k;  
21.     }  
22.     return sum;  
23. }  
24. ll dp[101010];  
25. int main()  
26. {  
27.     while(cin>>n>>k)  
28.     {  
29.         memset(dp,0,sizeof(dp));  
30.         memset(c,0,sizeof(c));  
31.         for(int i=1;i<=n;i++)  
32.         {  
33.             scanf("%d",&x);  
34.             dp[i]=dp[i-1];  
35.             if(x>=k)dp[i]++;  
36.         }  
37.         ll ans=0;  
38.         add(n+1,1);  
39.         for(int i=1;i<=n;i++)  
40.         {  
41.             ans+=read(2*dp[i]-i +n+1);  
42.             add(2*dp[i]-i +n+1,1);  
43.         }  
44.         printf("%lld\n",ans);  
45.     }  
46. }