POJ 2796 Feel Good

传送门

Time Limit: 3000MS　　Memory Limit: 65536K

Case Time Limit: 1000MS　　Special Judge

 

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

---

Solution

单调栈

---

Implementation

#include <cstdio>
#include <stack>
#include <iostream>
using namespace std;
typedef long long LL;

const int N(1e5+);

int h[N], L[N], R[N], st[N];
LL s[N];

int main(){
    int n;
    scanf("%d", &n);
    for(int i=; i<=n; i++) scanf("%d", h+i);
    for(int i=; i<=n; i++) s[i]=s[i-]+h[i];
    // (L[i], R[i]]
    int t=-;    //top of stack
    for(int i=; i<=n; i++){
        for(; ~t && h[st[t]]>=h[i]; t--);
        L[i]=~t?st[t]:;    //error-prone
        st[++t]=i;
    }
    t=-;
    for(int i=n; i; i--){
        for(; ~t && h[st[t]]>=h[i]; t--);
        R[i]=~t?st[t]-:n;
        st[++t]=i;
    }
    LL res=-;    //error-prone
    int l, r;
    for(int i=; i<=n; i++){
        LL t=h[i]*(s[R[i]]-s[L[i]]);
        if(t>res){
            res=t;
            l=L[i]+, r=R[i];
        }
    }
    printf("%lld\n%d %d\n", res, l, r);
    return ;
}

代码里标error-prone的地方都是坑，请特别注意。

尤其 res应初始化成-1，因为"A new idea Bill has recently developed assigns a non-negative integer value to each day of human life."