[ABC261D] Flipping and Bonus

Problem Statement

Takahashi will toss a coin $N$ times. He also has a counter, which initially shows $0$.

Depending on the result of the $i$-th coin toss, the following happens:

• If it heads: Takahashi increases the counter's value by $1$ and receives $X_i$ yen (Japanese currency).
• If it tails: he resets the counter's value to $0$, without receiving money.

Additionally, there are $M$ kinds of streak bonuses. The $i$-th kind of streak bonus awards $Y_i$ yen each time the counter shows $C_i$.

Find the maximum amount of money that Takahashi can receive.

Constraints

• $1\leq M\leq N\leq 5000$
• $1\leq X_i\leq 10^9$
• $1\leq C_i\leq N$
• $1\leq Y_i\leq 10^9$
• $C_1,C_2,\ldots,C_M$ are all different.
• All values in input are integers.

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Input

Input is given from Standard Input in the following format:

$N$ $M$

\(X_1\) \(X_2\) \(\ldots\) \(X_N\)

\(C_1\) \(Y_1\)

\(C_2\) \(Y_2\)

\(\vdots\)

\(C_M\) \(Y_M\)

Output

Print the maximum amount of money that Takahashi can receive, as an integer.

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Sample Input 1

6 3
2 7 1 8 2 8
2 10
3 1
5 5

Sample Output 1

48

If he gets head, head, tail, head, head, head, in this order, the following amounts of money are awarded.

• In the $1$-st coin toss, the coin heads. Change the counter's value from $0$ to $1$ and receive $2$ yen.
• In the $2$-nd coin toss, the coin heads. Change the counter's value from $1$ to $2$ and receive $7$ yen. Additionally, get $10$ yen as a streak bonus.
• In the $3$-rd coin toss, the coin tails. Change the counter's value from $2$ to $0$.
• In the $4$-th coin toss, the coin heads. Change the counter's value from $0$ to $1$ and receive $8$ yen.
• In the $5$-th coin toss, the coin heads. Change the counter's value from $1$ to $2$ and receive $2$ yen. Additionally, get $10$ yen as a streak bonus.
• In the $6$-th coin toss, the coin heads. Change the counter's value from $2$ to $3$ and receive $8$ yen. Additionally, get $1$ yen as a streak bonus.

In this case, Takahashi receives $2+(7+10)+0+8+(2+10)+(8+1)=48$ yen in total, which is the maximum possible.
Note that streak bonuses are awarded any number of times each time the counter shows $C_i$.
As a side note, if he gets head in all $6$ coin tosses, he only receives $2+(7+10)+(1+1)+8+(2+5)+8=44$ yen, which is not the maximum.

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Sample Input 2

3 2
1000000000 1000000000 1000000000
1 1000000000
3 1000000000

定义\(dp_{i,j}\)为扔了 \(i\) 次硬币 counter 为 \(j\) 的最大收钱数。
设 counter 达到 \(k\) 时收到 \(t_k\) 日元。
如果 \(j\ne 0\)，那么 \(dp_{i,j}=dp_{i-1,j-1}+X_i+t_j\)
如果 \(j=0\) 为 0，\(dp_{i,j}=\max\limits_{j=0}^{n-1}dp_{i-1,j}\)

#include<bits/stdc++.h>
using namespace std;
const int N=5005;
int x[N],c,y,t[N],n,m;
long long dp[N][N],ans;
int main()
{
	memset(dp,-0x7f,sizeof(dp));
	dp[0][0]=0;
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;i++)
		scanf("%d",x+i);
	for(int i=1;i<=m;i++)
		scanf("%d%d",&c,&y),t[c]+=y;
	for(int i=1;i<=n;i++)
	{
		dp[i][0]=dp[i-1][0];
		for(int j=1;j<=n;j++)
			dp[i][j]=dp[i-1][j-1]+t[j]+x[i],dp[i][0]=max(dp[i][0],dp[i-1][j-1]),ans=max(ans,dp[i][j]);
	}
	printf("%lld",ans);
}