[poj2796]Feel Good（单调栈）

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 10 ⁶ - 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

【题解】

通过这道题可以了解单调栈的一种功能。

这道题有两个限制条件：要求区间的总和，而都是正数，当然希望区间越长越好；但是还有最小值的限制，所以我们可以想到以一个点为中心，假定这个点是区间的最小值，然后向两边扩展，求出这个区间（当然是要最长的）。即f[i].l，f[i].r分别表示以a[i]为最小值的最长区间的左端点和右端点（也就是说最多到哪里元素都比a[i]大）。

元素进入单调栈的时候，先把栈内所有的大于它的元素都弹出，然后再将它压入（维护由栈顶向栈底递减的单调栈）。由单调栈的单调性可知，元素出栈时一定比将要进栈的元素大，将要进栈的元素一定比将要出栈的元素小，出栈后栈顶的元素一定比刚刚出栈的元素小（但不一定要比马上要入栈的元素小，还有可能要继续出栈），且栈底元素的序号一定比栈顶元素小。所以但凡遇到元素出栈（在操作过程中元素出栈往往由新元素入栈引起），将要进栈的元素的l向左延伸，出栈后栈顶元素的r向右延伸，且完全出栈后（栈顶所有比当前元素大的元素都已弹出），栈顶元素的r向右延伸。

注意最后要把栈里的元素全部弹出，并且也要进行上面的操作。

之后枚举每一个f，用前缀和求出区间和，更新答案即可。

【代码】

#include
#include
#include
#define ll long long
using namespace std;
struct hp{
	ll l,r;
}f[100005];
int n,ansl,ansr,tmp;
ll sum,Max;
ll a[100005],strack[100005];
ll s[100005];
inline void push(ll x){
	if (x==1){
		strack[++tmp]=x;
		return;
	}
	while (a[strack[tmp]]>=a[x]&&tmp>0){
		f[x].l=f[strack[tmp]].l;//将要进栈的l向左延伸
		f[strack[tmp-1]].r=f[strack[tmp]].r;//出栈后栈顶元素的r向右延伸
		--tmp;
	}
	f[strack[tmp]].r=f[x].r;//完全出栈后栈顶元素的r向右延伸
	strack[++tmp]=x;
}
inline void pop(){
	f[strack[tmp-1]].r=f[strack[tmp]].r;
	tmp--;
}
int main(){
	freopen("a.in","r",stdin);
	freopen("a.out","w",stdout);
	scanf("%d",&n);
	for (int i=1;i<=n;++i)
	  scanf("%d",&a[i]),s[i]=s[i-1]+a[i],f[i].l=f[i].r=i;
	for (int i=1;i<=n;++i)
	  push(i);
	while (tmp>0) pop();
	for (int i=1;i<=n;++i){
		sum=s[f[i].r]-s[f[i].l-1];
		if (Max<=sum*a[i]){
			Max=sum*a[i];
			ansl=f[i].l,ansr=f[i].r;
		}
	}
	printf("%I64d\n%d %d\n",Max,ansl,ansr);
}