[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.

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.

Sample Input 1

6 3

2 7 1 8 2 8

2 10

3 1

5 5

Sample Output 1

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.

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);

}