CodeForces 1197D Yet Another Subarray Problem

Time limit 2000 ms

Memory limit 262144 kB

Source Educational Codeforces Round 69 (Rated for Div. 2)

Tags dp greedy math *1900

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官方题解

At first let's solve this problem when m=1$m=1$ and k=0$k=0$ (it is the problem of finding subarray with maximum sum). For each position from 1$1$ to n$n$ we want to know the value of maxli=max1≤j≤i+1sum(j,i)$max{l}_i=\underset{1\le j\le i+1}{max}sum(j,i)$, where sum(l,r)=∑k=lk≤rak$sum(l,r)=\sum _{k=l}^{k\le r}{a}_k$, and sum(x+1,x)=0$sum(x+1,x)=0$.

We will calculate it the following way. maxli$max{l}_i$ will be the maximum of two values:

• 0$0$ (because we can take segments of length 0$0$);
• ai+maxli−1$a_i+max{l}_{i-1}$.

The maximum sum of some subarray is equal to max1≤i≤nmaxli$\underset{1\le i\le n}{max}max{l}_i$.

So, now we can calculate the values of besti=max0≤len,i−len⋅m≥0(sum(i−len⋅m+1,i)−len∗k)$bes{t}_i=\underset{0\le len,i-len\cdot m\ge 0}{max}(sum(i-len\cdot m+1,i)-len\ast k)$ the same way.

besti$bes{t}_i$ is the maximum of two values:

• 0;
• sum(i−m+1,i)−k+besti−m$sum(i-m+1,i)-k+bes{t}_{i-m}$.

After calculating all values besti$bes{t}_i$ we can easily solve this problem. At first, let's iterate over the elements besti$bes{t}_i$. When we fix some element besti$bes{t}_i$, lets iterate over the value len=1,2,…,m$len=1,2,\dots ,m$ and update the answer with value besti+sum(i−len,i−1)−k$bes{t}_i+sum(i-len,i-1)-k$.

源代码

#include<stdio.h>
#include<algorithm>

int n,m,k;
long long a[300010];
long long dp[300010],ans;
int main()
{
	//freopen("test.in","r",stdin);
	scanf("%d%d%d",&n,&m,&k);
	for(int i=1;i<=n;i++) scanf("%lld",a+i),a[i]+=a[i-1];
	for(int i=1;i<=n;i++)
	{
		for(int j=i;j+m>=i;j--)
			dp[i]=std::max(dp[i],a[i]-a[j]);
		dp[i]-=k;
		dp[i]=std::max(0LL,dp[i]);
		if(i>m) dp[i]=std::max(dp[i],dp[i-m]+a[i]-a[i-m]-k);
		ans=std::max(dp[i],ans);
	}
	printf("%lld
",ans);
	return 0;
}