HDU 2993 MAX Average Problem（斜率优化DP）

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

题目大意：给定一个长度为n（最长为10^5）的正整数序列，求出连续的最短为k的子序列平均值的最大值。

Sample Input

10 6

6 4 2 10 3 8 5 9 4 1

Sample Output

6.50

分析：斜率优化DP，要认真看

代码如下：

 1 # include<iostream>
 2 # include<cstdio>
 3 # include<cstring>
 4 # include<algorithm>
 5 
 6 using namespace std;
 7 
 8 const int maxn = 100010;
 9 double a[maxn], sum[maxn];
10 int n,k;
11 int q[maxn],head,tail;
12 
13 //读入优化，否则超时
14 int GetInt()
15 {
16     char ch=getchar();
17     while(ch<'0' || ch>'9')
18         ch = getchar();
19     int num = 0;
20     while(ch >= '0' && ch<='9')
21     {
22         num = num*10 + ch - '0';
23         ch = getchar();
24     }
25     return num;
26 }
27 
28 void DP()
29 {
30     head = tail =0;
31     double ans = -1;
32     for(int i=k; i<=n; i++)
33     {
34         int j = i-k;
35         //维护下凸
36         while(tail - head >=2)
37         {
38             double x1 = j - q[tail-1];
39             double y1 = sum[j] - sum[q[tail-1]];
40             double x2 = q[tail-1] - q[tail-2];
41             double y2 = sum[q[tail-1]] - sum[q[tail-2]];
42             if(x1 * y2 - y1 *x2 >= 0)
43                 tail--;
44             else
45                 break;
46         }
47         q[tail++] = j;
48         //寻找最优解并删除无用元素
49         while(tail - head >=2)
50         {
51             double x1 = i - q[head];
52             double y1 = sum[i] - sum[q[head]];
53             double x2 = i - q[head+1];
54             double y2 = sum[i] - sum[q[head+1]];
55             if(x1*y2 - y1*x2 >= 0)
56                 head ++;
57             else
58                 break;
59         }
60         double tmp = (sum[i] - sum[q[head]])/(i-q[head]);
61         ans = max(ans, tmp);
62     }
63     printf("%.2lf
",ans);
64 }
65 
66 int main()
67 {
68     //freopen("in.txt","r",stdin);
69     while(scanf("%d%d",&n,&k)!=EOF)
70     {
71         sum[0] = 0;
72         for(int i=1; i<=n; i++)
73         {
74             a[i] = GetInt();
75             sum[i] = sum[i-1] + a[i];
76         }
77         DP();
78     }
79     return 0;
80 }