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 = 100010;

int sum[MAXN];
int n, k;

int readint() {
    char c = getchar();
    while(!isdigit(c)) c = getchar();
    int res = 0;
    while(isdigit(c)) res = res * 10 + c - '0', 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 = 0;
    head = 0; tail = -1;
    for(int i = k; i <= n; ++i) {
        p = Point(i - k, sum[i - k]);
        while(head < tail && cross(que[tail - 1], que[tail], p) <= 0) --tail;
        que[++tail] = p;

        p = Point(i, sum[i]);
        while(head < tail && cross(que[head], que[head + 1], p) >= 0) ++head;
        res = max(res, (p - que[head]).slope());
    }
    return res;
}

int main() {
    while(scanf("%d%d", &n, &k) != EOF) {
        for(int i = 1; i <= n; ++i) sum[i] = sum[i - 1] + readint();
        printf("%.2f\n", solve());
    }
}