POJ 1742 coins (monotone queue solution) [suggestions collection]

Hello everyone , I meet you again , I'm the king of the whole stack .

Blog .

Plus, if the volume is equal to the value, it's better to do . Only j %v[ i ] The rest of the same ability can be handled at the same time （j It refers to the value of a certain volume ）, When calculating a number , Just calculate the same remainder as before （ Within the number limit ） Is there a true（ It's full ） You can .

Code ：

#include<iostream>
#include<sstream>
#include<map>
#include<cmath>
#include<fstream>
#include<queue>
#include<vector>
#include<sstream>
#include<cstring>
#include<cstdio>
#include<stack>
#include<bitset>
#include<ctime>
#include<string>
#include<cctype>
#include<iomanip>
#include<algorithm>
using namespace std  ;
#define INT long long int
#define L(x)  (x * 2)
#define R(x)  (x * 2 + 1)
const int INF = 0x3f3f3f3f ;
const double esp = 0.0000000001 ;
const double PI = acos(-1.0) ;
const int mod = 1000000007 ;
const int MY = (1<<5) + 5 ;
const int MX = 100010 + 5 ;
int n ,W ,ans ;
int v[MX] ,num[MX] ;
bool deq[MX] ,dp[MX] ;
void input()
{
    memset(dp ,false ,sizeof(dp)) ;
    for(int i = 1 ;i <= n ; ++i)
        scanf("%d" ,&v[i]) ;
    for(int i = 1 ;i <= n ; ++i)
        scanf("%d" ,&num[i]) ;
}
void DP(int v ,int num)
{
    if(!num || !v)  return ;
    if(num == 1) // 01  knapsack 
    {
        for(int i = W ;i >= v ; --i)
           if(!dp[i] && dp[i-v])
               dp[i] = true ,ans++ ;
    }
    else if(num * v >= W) //  It's all backpacks 
    {
        for(int i = v ;i <= W ; ++i)
          if(!dp[i] && dp[i-v])
              dp[i] = true ,ans++ ;
    }
    else
    {
        num = min(num ,W/v) ;
        for(int a = 0 ;a < v ; ++a)  //  Similarly, the remainder is processed together 
        {
            int front =0 ,end = 0 ,sum = 0 ;
            for(int j = a ;j <= W ; j += v)
            {
                if(end - front-1 == num)  //  Remove obsolete elements  , Due to the most choices num[i]  individual 
                    sum -= deq[front++] ;
                deq[end++] = dp[j] ;  //  Deposit in 
                sum += dp[j] ;
                if(!dp[j] && sum)
                    dp[j] = true ,ans++ ;

            }
        }
    }
}
int main()
{
    //freopen("input.txt" ,"r" ,stdin) ;
    while(scanf("%d%d" ,&n ,&W) ,n+W)
    {
        input() ;
        dp[0] = true ;
        ans = 0 ;
        for(int i = 1 ;i <= n ; ++i)
             DP(v[i] ,num[i]) ;
        printf("%d\n" ,ans) ;
    }
    return 0 ;
}

Copy

Publisher ： Full stack programmer stack length , Reprint please indicate the source ：https://javaforall.cn/116515.html Link to the original text ：https://javaforall.cn