[ACM] POJ 2796 Feel Good （求序列中子序列的和与子序列中的最小数最大乘积）

Feel Good

Description

Bill is developing a new mathematical theory for human emotions. His recent investigations are dedicated to studying how good or bad days influent people‘s memories about some period of life.

A new idea Bill has recently developed assigns a non-negative integer value to each day of human life.

Bill calls this value the emotional value of the day. The greater the emotional value is, the better the daywas. Bill suggests that the value of some period of human life is proportional to the sum of the emotional values of the days in the given period, multiplied
by the smallest emotional value of the day in it. This schema reflects that good on average period can be greatly spoiled by one very bad day.

Now Bill is planning to investigate his own life and find the period of his life that had the greatest value. Help him to do so.

Input

The first line of the input contains n - the number of days of Bill‘s life he is planning to investigate(1 <= n <= 100 000). The rest of the file contains n integer numbers a1, a2, ... an ranging from 0 to 10⁶ - the emotional values of the days.
Numbers are separated by spaces and/or line breaks.

Output

Print the greatest value of some period of Bill‘s life in the first line. And on the second line print two numbers l and r such that the period from l-th to r-th day of Bill‘s life(inclusive) has the greatest possible value. If there are multiple periods with
the greatest possible value,then print any one of them.

Sample Input

6
3 1 6 4 5 2

Sample Output

60
3 5

Source

Northeastern Europe 2005

解题思路：

给定一序列，求一个子序列要求子序列中的和与子序列中的最小数乘积最大，输出最大乘积以及子序列的起始位置。

我们可以转化为求序列中的每一个数所属于哪个区间，保证该数在这个区间内是最小的，并且这个区间长度最大（比如  3 4 2  6,     2所属区间可以为（4,2）也可以为（4,2,6），我们要求最大的(3,4,2,6)，因为要保证子序列的和最大）。

这样求出每个数的所属区间，比较一下就可以了。传统的求每个数所属区间时间复杂度高O（n2)，可以采用动规的思想，也可以叫回溯，也和KMP的next数组思想有点像

求第i个数所属区间的左端点L[i]时,L[i]肯定小于等于i，所以把i从第i-1个数（i前面的）开始比较，如果第i-1个数比第i个数小，那么肯定L[i] = i+1，第i-1个数小于第i个数，肯定不会在第i个数所属的区间，如果第i-1个数比第i个数大，那么就用第i个数与  L[ i -1] -1 （第i-1个数所属区间左端点的前一个数)进行比较。

求右端点的思想和求左端点的思想是一样的，逆向求。

代码：

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
const int maxn=100010;
int s[maxn];//存放第i个元素在某个区间内为最小值的区间左端点
int e[maxn];//存放第i个元素在某个区间内为最小值的区间右端点
int node[maxn];//存放输入元素
long long ans;//输出结果
long long sum[maxn];//存放第1到i个元素的和
int S=1,E=1;//输出所求区间的起始位置

int main()
{
    int n;cin>>n;
    sum[0]=0;//以下三行初始化为后面做准备
    node[0]=-1;
    node[n+1]=-1;
    for(int i=1;i<=n;i++)//求第1到i个元素和，并且求出第i个元素所属区间的左端点
    {
        scanf("%d",&node[i]);
        sum[i]=sum[i-1]+node[i];//求和
        for(int j=i-1;j>=0;)//第i个元素从它前面的第一个元素比较
        {
            if(node[i]>node[j])
            {
                s[i]=j+1;//肯定是j+1，因为第j个元素肯定不在第i个元素所属区间内，得保证第i个元素在所属区间内为最小值
                break;
            }
            else
                j=s[j]-1;//回溯，从第j个元素所属区间的前面第一个开始比较，如 1 5 4 2， 2<4,4所属区间为（5,4），s[j]=2,所以与第一个元素1比较
        }
    }
    for(int i=n;i>=1;i--)//逆向求第i个元素所属区间的右端点
    {
        for(int j=i+1;j<=n+1;)//从i后面的第一个元素比较。
        {
            if(node[i]>node[j])//原理同上
            {
                e[i]=j-1;
                break;
            }
            else
                j=e[j]+1;
        }
    }
    long long temp=0;
    ans=-1;
    for(int i=1;i<=n;i++)
    {
        temp=(sum[e[i]]-sum[s[i]-1])*node[i];
        if(temp>ans)
        {
            ans=temp;
            S=s[i];
            E=e[i];
        }
    }
    cout<<ans<<endl;
    cout<<S<<" "<<E<<endl;
    return 0;
}