ABC 261.D - Flipping and Bonus ( DP )

Problem statement
Flip a coin N Time , There is a counter , Initially displayed as 0.

According to the first i The result of the coin toss , The following will happen :

If it is positive ： Increase the value of the counter 1 And get the Xi;
If it's the opposite ： Set the value of the counter to 0（ No addition Xi）.
Besides , Yes M Winning streak bonus .

Ask for the maximum amount you can receive .

Sample Input 1 Copy

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

Sample Output 1 Copy

48

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

• 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 22 to 3 and receive 8 yen. Additionally, get 1 yen as a streak bonus.

max operation ：2+(7+10)+0+8+(2+10)+(8+1)=48

Note the length of the two-dimensional array ！

//#include<bits/stdc++.h>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<string>
#include<cstdio>
#include<map>
#include<vector>
#include<set>
#include<queue>
#include<deque>
using namespace std;
typedef long long LL;
typedef pair<int,int> PII;
const int N=200200;
const int M=5002;
const int mod=998244353;
LL a[N],f[M][M];
LL mp[N];
int b[N];
bool vis[N];
int dx[]={-1,0,0,1},dy[]={0,1,-1,0};
string s[N];
int main()
{
    cin.tie(0); cout.tie(0); ios::sync_with_stdio(false);
    int T=1;
    //cin>>T;
    while(T--)
    {
        int n,m;
        cin>>n>>m;
        for(int i=1;i<=n;i++)
            cin>>a[i];
        for(int i=1;i<=m;i++)
        {
            LL x,y;
            cin>>x>>y;
            mp[x]=y;
        }
        //f[i][j] Co throw i Times and the counter is j
        for(int i=1;i<=n;i++)// Throw it all i The second coin 
        {
            for(int j=1;j<=i;j++)// The counter for j
            {
                // Let the coin thrown at present experience the value at what position 
                f[i][j]=f[i-1][j-1]+a[i]+mp[j];
            }
            f[i][0]=0;
            for(int k=1;k<i;k++)
            {
                // Record the maximum value that can be obtained under the operation of the previous line 
                // Put it in the 0 The position of is to meet the above when the counter is 1 When 
                f[i][0]=max(f[i][0],f[i-1][k]);
            }
        }
        /*for(int i=0;i<=n;i++)
        {
            for(int j=0;j<=n;j++)
            {
                cout<<f[i][j]<<" ";
            }
            cout<<endl;
        }*/
        LL maxn=0;
        for(int i=1;i<=n;i++)
            maxn=max(maxn,f[n][i]);
        cout<<maxn<<endl;
    }
    return 0;
}