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：

 # include<iostream>
 # include<cstdio>
 # include<cstring>
 # include<algorithm>

 using namespace std;

 const int maxn = ;
 double a[maxn], sum[maxn];
 int n,k;
 int q[maxn],head,tail;

 //读入优化，否则超时
 int GetInt()
 {
     char ch=getchar();
     while(ch<'' || ch>'')
         ch = getchar();
     int num = ;
     while(ch >= '' && ch<='')
     {
         num = num* + ch - '';
         ch = getchar();
     }
     return num;
 }

 void DP()
 {
     head = tail =;
     double ans = -;
     for(int i=k; i<=n; i++)
     {
         int j = i-k;
         //维护下凸
         while(tail - head >=)
         {
             double x1 = j - q[tail-];
             double y1 = sum[j] - sum[q[tail-]];
             double x2 = q[tail-] - q[tail-];
             double y2 = sum[q[tail-]] - sum[q[tail-]];
             if(x1 * y2 - y1 *x2 >= )
                 tail--;
             else
                 break;
         }
         q[tail++] = j;
         //寻找最优解并删除无用元素
         while(tail - head >=)
         {
             double x1 = i - q[head];
             double y1 = sum[i] - sum[q[head]];
             double x2 = i - q[head+];
             double y2 = sum[i] - sum[q[head+]];
             if(x1*y2 - y1*x2 >= )
                 head ++;
             else
                 break;
         }
         double tmp = (sum[i] - sum[q[head]])/(i-q[head]);
         ans = max(ans, tmp);
     }
     printf("%.2lf\n",ans);
 }

 int main()
 {
     //freopen("in.txt","r",stdin);
     while(scanf("%d%d",&n,&k)!=EOF)
     {
         sum[] = ;
         for(int i=; i<=n; i++)
         {
             a[i] = GetInt();
             sum[i] = sum[i-] + a[i];
         }
         DP();
     }
     return ;
 }