POJ2796 Feel Good

题目

题目

思路

2遍单调栈算贡献的左右边界，用前缀和预处理区间和，计算最优解。
code:

#include<iostream>
#include<stack>
#include<cstdio>
using namespace std;
stack<long long> u,u2;
long long n,a[100005],s[100005],l[100005],r[100005],ans,sl=1,sr=1,i;
int main()
{
 scanf("%lld",&n);
 for (i=1;i<=n;i++)
 {
  scanf("%lld",&a[i]);
  s[i]=s[i-1]+a[i];
 }
 for (i=1;i<=n;i++)
 {
  while (u.size()&&a[u.top()]>=a[i]) u.pop();
  l[i]=(u.size()==0?1:u.top()+1);
  u.push(i);
 }
 for (i=n;i>=1;i--)
 {
  while (u2.size()&&a[u2.top()]>=a[i]) u2.pop();
  r[i]=(u2.size()==0?n:u2.top()-1);
  u2.push(i);
  if (ans<(s[r[i]]-s[l[i]-1])*a[i])
  {
   ans=(s[r[i]]-s[l[i]-1])*a[i];
   sl=l[i],sr=r[i];
  }
 }
 printf("%lld\n%lld %lld\n",ans,sl,sr);
 return 0;
}

标签：Good,Feel,top,u2,100005,ans,size,POJ2796,lld
来源： https://blog.csdn.net/weixin_49843717/article/details/114433508