F2. Pictures with Kittens (hard version)
time limit per test
2.5 seconds
memory limit per test
512 megabytes
input
standard input
output
standard output
The only difference between easy and hard versions is the constraints.
Vova likes pictures with kittens. The news feed in the social network he uses can be represented as an array of nn consecutive pictures (with kittens, of course). Vova likes all these pictures, but some are more beautiful than the others: the ii-th picture has beauty aiai.
Vova wants to repost exactly xx pictures in such a way that:
- each segment of the news feed of at least k consecutive pictures has at least one picture reposted by Vova;
- the sum of beauty values of reposted pictures is maximum possible.
For example, if k=1 then Vova has to repost all the pictures in the news feed. If k=2 then Vova can skip some pictures, but between every pair of consecutive pictures Vova has to repost at least one of them.
Your task is to calculate the maximum possible sum of values of reposted pictures if Vova follows conditions described above, or say that there is no way to satisfy all conditions.
Input
The first line of the input contains three integers n,k and x (1≤k,x≤n≤5000) — the number of pictures in the news feed, the minimum length of segment with at least one repost in it and the number of pictures Vova is ready to repost.
The second line of the input contains nn integers a1,a2,…,an (1≤ai≤109), where aiai is the beauty of the i-th picture.
Output
Print -1 if there is no way to repost some pictures to satisfy all the conditions in the problem statement.
Otherwise print one integer — the maximum sum of values of reposted pictures if Vova follows conditions described in the problem statement.
Examples
input
Copy
5 2 3
5 1 3 10 1
output
Copy
18
input
Copy
6 1 5
10 30 30 70 10 10
output
Copy
-1
input
4 3 1
1 100 1 1
output
100题意：
满足任意连续的k个数至少有一个选中，问选中x个的最大值，如果不存在输出-1，否则输出最大值
分析：
用单调队列维护区间k的最大值，减少一维
代码：
#include<bits/stdc++.h>
#define ll long long
#define mod 1000000007
#define inf 0x3f3f3f3f3f3f
using namespace std;
ll a[5105],q[5105];
ll dp[5100][5100];
int main()
{
ll i,j,n,k,x,l,r;
scanf("%lld%lld%lld",&n,&k,&x);
for(i=1;i<=n;i++)
scanf("%lld",&a[i]);
for(i=1;i<=n;i++)
dp[i][0]=-inf;
for(i=1;i<=x;i++)
{
l=r=0;
q[r++]=i-1;
for(j=i;j<=n;j++)
{
while(l<r&&q[l]<j-k)l++;
if(l<r)dp[j][i]=dp[q[l]][i-1]+a[j];
while(l<r&&dp[j][i-1]>=dp[q[r-1]][i-1])r--;
q[r++]=j;
}
}
ll maxn;
maxn=-inf;
for(i=n-k+1;i<=n;i++)
maxn=max(maxn,dp[i][x]);
if(maxn<=0)maxn=-1;
printf("%lld\n",maxn);
}