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）：

 #include <cstdio>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 #include <cctype>
 using namespace std;
 typedef long long LL;

 const int MAXN = ;

 int sum[MAXN];
 int n, k;

 int readint() {
     char c = getchar();
     while(!isdigit(c)) c = getchar();
     int res = ;
     while(isdigit(c)) res = res *  + c - '', c = getchar();
     return res;
 }

 struct Point {
     int x, y;
     Point() {}
     Point(int x, int y): x(x), y(y) {}
     Point operator - (const Point &rhs) const {
         return Point(x - rhs.x, y - rhs.y);
     }
     double slope() {
         return (double)y / x;
     }
 };

 LL cross(const Point &a, const Point &b) {
     return (LL)a.x * b.y - (LL)a.y * b.x;
 }

 LL cross(const Point &o, const Point &a, const Point &b) {
     return cross(a - o, b - o);
 }

 Point que[MAXN], p;
 int head, tail;

 double solve() {
     double res = ;
     head = ; tail = -;
     for(int i = k; i <= n; ++i) {
         p = Point(i - k, sum[i - k]);
         while(head < tail && cross(que[tail - ], que[tail], p) <= ) --tail;
         que[++tail] = p;

         p = Point(i, sum[i]);
         while(head < tail && cross(que[head], que[head + ], p) >= ) ++head;
         res = max(res, (p - que[head]).slope());
     }
     return res;
 }

 int main() {
     while(scanf("%d%d", &n, &k) != EOF) {
         for(int i = ; i <= n; ++i) sum[i] = sum[i - ] + readint();
         printf("%.2f\n", solve());
     }
 }