Codeforce1176F Destroy it!

Description

You are playing a computer card game called Splay the Sire. Currently you are struggling to defeat the final boss of the game.

The boss battle consists of n turns. During each turn, you will get several cards. Each card has two parameters: its cost ci and damage di. You may play some of your cards during each turn in some sequence (you choose the cards and the exact order they are played), as long as the total cost of the cards you play during the turn does not exceed 3. After playing some (possibly zero) cards, you end your turn, and all cards you didn't play are discarded. Note that you can use each card at most once.

Your character has also found an artifact that boosts the damage of some of your actions: every 10-th card you play deals double damage.

What is the maximum possible damage you can deal during n turns?

Input

The first line contains one integer n (1≤n≤2⋅\(10^{5}\)) — the number of turns.

Then n blocks of input follow, the i-th block representing the cards you get during the i-th turn.

Each block begins with a line containing one integer ki (1≤ki≤2⋅\(10^{5}\)) — the number of cards you get during i-th turn. Then ki lines follow, each containing two integers cj and dj (1≤cj≤3, 1≤dj≤\(10^{9}\)) — the parameters of the corresponding card.

It is guaranteed that \(\sum_{i=1}^{n}\)ki≤2⋅\(10^{5}\).

Output

Print one integer — the maximum damage you may deal.

Solution

就是一個01揹包加入了一些特殊的條件：
1.將這些物品分紅n塊，每塊的質量之和不能超過3
2.當選到第10的倍數個數時，該數貢獻翻倍
設f[i][j]表示作到第i個數，選了j個數的最大價值和
顯然\(n^{2}\)咱們沒法接受，因而咱們把j滾動
其餘和01揹包是同樣的

Code

#include <cstdio>
#include <algorithm>
#include <cstring>
#define N 200001
#define open(x) freopen(x".in","r",stdin);freopen(x".out","w",stdout);
using namespace std;
int n,i,j,k,a,b,t,ma,mi,m1[N][10],m2[N],m3[N];
long long ans,f[N][10];
int main()
{
    open("destroy");
    scanf("%d",&n);
    for (i=1;i<=n;i++)
    {
        scanf("%d",&t);
        for (j=1;j<=t;j++)
        {
            scanf("%d%d",&a,&b);
            if (a==1)
            {
                if (b>m1[i][1]) m1[i][3]=m1[i][2],m1[i][2]=m1[i][1],m1[i][1]=b;else 
                if (b>m1[i][2]) m1[i][3]=m1[i][2],m1[i][2]=b;else 
                if (b>m1[i][3]) m1[i][3]=b;
            }
            if (a==2) m2[i]=max(m2[i],b);
            if (a==3) m3[i]=max(m3[i],b);
        }
    }
    memset(f,128,sizeof(f));
    f[0][0]=0;
    for (i=1;i<=n;i++)
    {
        for (j=0;j<=9;j++) f[i][j]=f[i-1][j];
        for (k=0;k<=9;k++)
        {
            if (f[i-1][k]>=0) 
            {
                if (m1[i][1])f[i][(k+1)%10]=max(f[i][(k+1)%10],f[i-1][k]+m1[i][1]*(k==9?2:1));
                if (m1[i][1]&&m1[i][2]) f[i][(k+2)%10]=max(f[i][(k+2)%10],f[i-1][k]+m1[i][1]*(k>=8?2:1)+m1[i][2]);
                if (m2[i]) f[i][(k+1)%10]=max(f[i][(k+1)%10],f[i-1][k]+m2[i]*(k==9?2:1));
                if (m1[i][1]&&m1[i][2]&&m1[i][3]) f[i][(k+3)%10]=max(f[i][(k+3)%10],f[i-1][k]+m1[i][1]*(k>=7?2:1)+m1[i][2]+m1[i][3]);
                if (m1[i][1]&&m2[i]) 
                {
                    ma=max(m1[i][1],m2[i]);mi=min(m1[i][1],m2[i]);
                    f[i][(k+2)%10]=max(f[i][(k+2)%10],f[i-1][k]+ma*(k>=8?2:1)+mi);
                }
                if (m3[i]) f[i][(k+1)%10]=max(f[i][(k+1)%10],f[i-1][k]+m3[i]*(k==9?2:1));
            }
        }
    }
    for (i=0;i<=9;i++) ans=max(ans,f[n][i]);
    printf("%lld",ans);
    return 0;
}