HDU_3415_单调队列

Posted 2020-08-11 冷暖知不知

http://acm.hdu.edu.cn/showproblem.php?pid=3415

 

初探单调队列，需要注意的是每次i维护的是i-1。

 

#include<iostream>
#include<cstring>
#include<cstdio>
#include<deque>
#define INF 0x3f3f3f3f
using namespace std;
int a[200005],sum[200005];
deque<int> q;
int main()
{
    int T,n,k;
    scanf("%d",&T);
    while(T--)
    {
        q.clear();
        sum[0] = 0;
        scanf("%d%d",&n,&k);
        for(int i = 1;i <= n;i++)
        {
            scanf("%d",&a[i]);
            sum[i] = sum[i-1]+a[i];
        }
        for(int i = 1;i < n;i++)    sum[i+n] = sum[i+n-1]+a[i];
        int ans = -INF,left,right;
        int endd = 2*n;
        for(int i = 1;i < endd;i++)
        {
            while(!q.empty() && sum[i-1] < sum[q.back()])   q.pop_back();
            if(!q.empty() && q.front() < i-k)   q.pop_front();
            q.push_back(i-1);
            if(sum[i]-sum[q.front()] > ans)
            {
                ans = sum[i]-sum[q.front()];
                left = q.front()+1;
                right = i;
            }
        }
        if(right > n)   right -= n;
        printf("%d %d %d\n",ans,left,right);
    }
    return 0;
}