HDU 2993 MAX Average Problem（斜率优化）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2993

Problem Description

Consider a simple sequence which only contains positive integers as a1, a2 ... an, and a number k. Define ave(i,j) as the average value of the sub sequence ai ... aj, i<=j. Let’s calculate max(ave(i,j)), 1<=i<=j-k+1<=n.

 

Input

There multiple test cases in the input, each test case contains two lines.
The first line has two integers, N and k (k<=N<=10^5).
The second line has N integers, a1, a2 ... an. All numbers are ranged in [1, 2000].

 

Output

For every test case, output one single line contains a real number, which is mentioned in the description, accurate to 0.01.

 

题目大意：给n个数和k，求数的个数大于等于k的子段的最大平均值。

思路：可以去看IOI国家集训队论文：《浅谈数形结合思想在信息学竞赛中的应用》——周源

也可以直接看这个http://blog.sina.com.cn/s/blog_ad1f8960010174el.html

 

PS：这就是传说中的来自数据组数的恶意吗，看上去似乎有100组大数据的感觉……G++超时的可以尝试用C++交，HDU的C++读入比G++快，而且优化的程度也不同。

 

代码（C++ 500MS/G++ 906MS）：

 1 #include <cstdio>

 2 #include <cstring>

 3 #include <iostream>

 4 #include <algorithm>

 5 #include <cctype>

 6 using namespace std;

 7 typedef long long LL;

 8 

 9 const int MAXN = 100010;

10 

11 int sum[MAXN];

12 int n, k;

13 

14 int readint() {

15     char c = getchar();

16     while(!isdigit(c)) c = getchar();

17     int res = 0;

18     while(isdigit(c)) res = res * 10 + c - '0', c = getchar();

19     return res;

20 }

21 

22 struct Point {

23     int x, y;

24     Point() {}

25     Point(int x, int y): x(x), y(y) {}

26     Point operator - (const Point &rhs) const {

27         return Point(x - rhs.x, y - rhs.y);

28     }

29     double slope() {

30         return (double)y / x;

31     }

32 };

33 

34 LL cross(const Point &a, const Point &b) {

35     return (LL)a.x * b.y - (LL)a.y * b.x;

36 }

37 

38 LL cross(const Point &o, const Point &a, const Point &b) {

39     return cross(a - o, b - o);

40 }

41 

42 Point que[MAXN], p;

43 int head, tail;

44 

45 double solve() {

46     double res = 0;

47     head = 0; tail = -1;

48     for(int i = k; i <= n; ++i) {

49         p = Point(i - k, sum[i - k]);

50         while(head < tail && cross(que[tail - 1], que[tail], p) <= 0) --tail;

51         que[++tail] = p;

52 

53         p = Point(i, sum[i]);

54         while(head < tail && cross(que[head], que[head + 1], p) >= 0) ++head;

55         res = max(res, (p - que[head]).slope());

56     }

57     return res;

58 }

59 

60 int main() {

61     while(scanf("%d%d", &n, &k) != EOF) {

62         for(int i = 1; i <= n; ++i) sum[i] = sum[i - 1] + readint();

63         printf("%.2f\n", solve());

64     }

65 }

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