hdu 3732 Ahui Writes Word 【多重背包】、好题_rejected by hdu waf-优快云博客

本题描述了一个关于选择单词以最大化总价值的问题，在限定复杂度的情况下，通过转换算法思路从0-1背包问题优化到多重背包问题，实现了高效的求解。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Ahui Writes Word

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2562 Accepted Submission(s): 931

Problem Description

We all know that English is very important, so Ahui strive for this in order to learn more English words. To know that word has its value and complexity of writing (the length of each word does not exceed 10 by only lowercase letters), Ahui wrote the complexity of the total is less than or equal to C.
Question: the maximum value Ahui can get.
Note: input words will not be the same.

Input

The first line of each test case are two integer N , C, representing the number of Ahui’s words and the total complexity of written words. (1 ≤ N ≤ 100000, 1 ≤ C ≤ 10000)
Each of the next N line are a string and two integer, representing the word, the value(Vi ) and the complexity(Ci ). (0 ≤ Vi , Ci ≤ 10)

Output

Output the maximum value in a single line for each test case.

Sample Input
5 20
go 5 8
think 3 7
big 7 4
read 2 6
write 3 5

Sample Output
15

0-1背包超时、转多重背包、从0-1背包转多重背包

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
int dp[1000000];
int wupin[11][11];
int v,n;
void zoreonepack(int val,int cost)
{
    for(int i=v;i>=cost;i--)
    {
        if(dp[i-cost]+val>dp[i])
        {
            dp[i]=dp[i-cost]+val;
        }
    }
}
void completepack(int val,int cost)
{
    for(int i=cost;i<=v;i++)
    {
        dp[i]=max(dp[i],dp[i-cost]+val);
    }
}
void multipack(int val,int cost,int num)
{
    if(num*cost>=v)
    {
        completepack(val,cost);
    }
    else
    {
        int k=1;
         while(k<num)
         {
             zoreonepack(k*val,k*cost);
             num-=k;k+=k;
         }
         zoreonepack(num*val,num*cost);
    }
}
int main()
{
    while(~scanf("%d%d",&n,&v))
    {
        memset(dp,0,sizeof(dp));
        memset(wupin,0,sizeof(wupin));
        for(int i=0;i<n;i++)
        {
            char s[1000];
            int x,y;
            scanf("%s%d%d",s,&x,&y);
            wupin[x][y]++;
        }
        for(int i=0;i<11;i++)
        {
            for(int j=0;j<11;j++)
            {
                if(wupin[i][j]!=0)
                {
                     multipack(i,j,wupin[i][j]);
                }
            }
        }
        printf("%d\n",dp[v]);
    }
}