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 = 100010;
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<'0' || ch>'9')
        ch = getchar();
    int num = 0;
    while(ch >= '0' && ch<='9')
    {
        num = num*10 + ch - '0';
        ch = getchar();
    }
    return num;
}

void DP()
{
    head = tail =0;
    double ans = -1;
    for(int i=k; i<=n; i++)
    {
        int j = i-k;
        //维护下凸
        while(tail - head >=2)
        {
            double x1 = j - q[tail-1];
            double y1 = sum[j] - sum[q[tail-1]];
            double x2 = q[tail-1] - q[tail-2];
            double y2 = sum[q[tail-1]] - sum[q[tail-2]];
            if(x1 * y2 - y1 *x2 >= 0)
                tail--;
            else
                break;
        }
        q[tail++] = j;
        //寻找最优解并删除无用元素
        while(tail - head >=2)
        {
            double x1 = i - q[head];
            double y1 = sum[i] - sum[q[head]];
            double x2 = i - q[head+1];
            double y2 = sum[i] - sum[q[head+1]];
            if(x1*y2 - y1*x2 >= 0)
                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[0] = 0;
        for(int i=1; i<=n; i++)
        {
            a[i] = GetInt();
            sum[i] = sum[i-1] + a[i];
        }
        DP();
    }
    return 0;
}