POJ 2796 Feel Good（单调栈）

传送门

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

思路

定义一串数字，其子区间的值为子区间元素和乘以区间最小值，因此利用单调栈的思想，以每个当前值为区间最小值，向左向右拓展区间。如果当前元素大于栈顶元素，那么就不能往前拓展，

如果当前元素小于栈顶元素，这个时候就要把栈中的元素一个一个弹出来，直到当前元素大于栈顶元素，对于弹出来的元素，它扩展到当前元素就不能向后伸展下去了，因此对于弹出来的元素这个时候就可以计算左右端点形成区间和与最小值的乘积了，维护一个最大值即可。

 

#include<stdio.h>
#include<string.h>
typedef __int64 LL;
const int maxn = 1000005;
LL a[maxn],stack[maxn] = {0},left[maxn] = {0},sum[maxn] = {0};

int main()
{
	LL N,L = 0,R = 0,res = -1,tmp = 0;
	scanf("%I64d",&N);
	for (int i = 1;i <= N;i++)	scanf("%I64d",&a[i]),sum[i] = a[i] + sum[i-1];
	a[++N] = -1;    //确保栈中元素能被全部弹出
	int top = 0;
	for (int i = 1;i <= N;i++)
	{
		if (!top || a[i] > a[stack[top-1]])
		{
			stack[top++] = i;
			left[i] = i;
			continue;
		}

		if (a[i] == a[stack[top-1]])	continue;

		while (top > 0 && a[i] < a[stack[top-1]])
		{
			top--;
			tmp = a[stack[top]]*(sum[i-1] - sum[left[stack[top]] - 1]);
			if (tmp > res)	res = tmp,L = left[stack[top]],R = i-1;
		}
		tmp = stack[top];
		stack[top++] = i;
		left[i] = left[tmp];
	}
	printf("%I64d\n%I64d %I64d\n",res,L,R);
	return 0;
}