POJ - 2796 - Feel Good POJ 【单调栈】题解-蒲公英云

POJ - 2796 - Feel Good POJ 【单调栈】题解

1.题目

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

2.代码

#include<stdio.h>
#include<iostream>
#include<stack>
using namespace std;
typedef long long LL;
int main()
{ 
    int i,n,pos1,pos2; //pos1和pos2记录区间的开始和结束位置 
    //tmp为临时变量，记录区间内的和；top指向栈顶元素；ans为结果；sum为前缀和 
    LL tmp,top,ans,a[100010],sum[100010];
    stack<int> st; //单调栈，记录元素位置 
    while(~scanf("%d",&n))
    { 
        while(!st.empty()) st.pop(); //清空栈 
        sum[0]=0;
        for(i=1;i<=n;i++)
        { 
            scanf("%lld",&a[i]);
            sum[i]=sum[i-1]+a[i]; //计算前缀和 
        }            
        a[n+1]=-1; //将最后一个设为最小值，以最后让栈内元素全部出栈 
        ans=0;
        for(i=1;i<=n+1;i++)
        { 
            if(st.empty()||a[i]>=a[st.top()])
            {  //如果栈为空或入栈元素大于等于栈顶元素，则入栈 
                st.push(i);
            }
            else 
            { 
                while(!st.empty()&&a[i]<a[st.top()])
                {  //如果栈非空并且入栈元素小于栈顶元素，则将栈顶元素出栈 
                    top=st.top();
                    st.pop();                    
                    tmp=sum[i-1]-sum[top-1]; //计算区间内元素和 
                    tmp*=a[top]; //计算结果 
                    if(tmp>=ans) 
                    {  //更新最大值并记录位置 
                        ans=tmp;
                        pos1=top;
                        pos2=i;
                    }
                }
                st.push(top); //将最后一次出栈的栈顶元素入栈 
                a[top]=a[i]; //将其向左向右延伸并更新对应的值 
            }
        }
        printf("%lld\n",ans);
        printf("%d %d\n",pos1,pos2-1);
    }
    return 0;
}