Mixed backpack (monotone optimization queue multiple backpack) (template)

Brief

As the title, put up a bit of a mixed backpack optimal time

 Code area

#include<bits/stdc++.h>
using namespace std;
const int Max = 1e5+10;

int n, v;
int val[Max], vol[Max], num[Max];
int que[Max];
int dp[Max];

void work()
{
    memset(dp, 0, sizeof(dp));
    for (int i = 1; i <= n; i++)
    {
        if (num[i] == 1)                    //01背包
        {
            for (intV = J; J> = [I] Vol; J, ) 
                DP [J] = max (DP [J], DP [J - Vol [I]] + Val [I]);
             Continue ; 
        } 
        IF (Vol [I] * NUM [I]> = V)             // complete backpack, the more critical, because very time-consuming process queue monotone 
        {
             for ( int J = Vol [I]; J <= V; J ++ ) 
                DP [J ] = max (DP [J], DP [J - Vol [I]] + Val [I]);
             Continue ; 
        } 
        for ( int RES = 0 ; RES <Vol [I]; RES ++)                             // enumerator I 
        {
            int head = 0 , tail = - . 1 ;
             for ( int K = 0 ; K <= (V - RES) / Vol [I]; K ++)                 // Enumeration k ', i.e., a number of arithmetic progression, each updated with the number of the remainder
                 // because only the same number of remainders will interact 
            {
                 int value DP = [K * Vol [I] + RES] - Val K * [I];             // current value 

                IF (tail - head K ==)                                     // even if all the materials can not be exhausted from the que [head] transfer 
                    head ++ ;
                 the while (head <= tail && que [tail] <= value)                 // takes a maximum value
                    tail--;
                tail++;
                que[tail] = value;
                dp[k * vol[i] + res] = que[head] + k * val[i];
            }
        }
    }
}

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