HDU3732 01背包转化为多重背包

Ahui Writes Word

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2738    Accepted Submission(s): 971

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

    
    
    
    
 
     
 
      Hint 
     
 Input data is huge,please use “scanf(“%s”,s)” 
    
    
    
    
     
   
   
   
   

 

Author

Ahui

 

Source

ACMDIY第二届群赛

题意：读题可知，每个单词只能出现一次，每个单词对应有价值和复杂度，很容易就可以想到是01背包的问题，可是普通的01背包肯定会超时，N*C<=10^9，数据量太大，但是每个单词的复杂度和价值都是介于0~10之间的，所以可以通过他们把01背包转化为多重背包。

#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <string.h>
#include <queue>
#include <string>
#include <stdlib.h>
#include <vector>
#include <map>
#include <set>
const int MAX = 0x3f3f3f3f;
const int MIN = -1<<28;
const int INF = 0x3f3f3f3f;
#define L(n,m) memset(n,m,sizeof(n);
#define M(i,n,m) for(int i = n;i < m;i ++)
using namespace std;

int n,C;
int a[11][11];
int dp[10010];
char s[10];

inline void ZERO_ONE_PACK(int value,int weight)
{
    for(int i = C;i >= weight;i --)
        dp[i] = max(dp[i],dp[i - weight] + value);
}

inline void COMPLETE_PACK(int value,int weight)
{
    for(int i = weight;i <= C;i ++)
        dp[i] = max(dp[i],dp[i - weight] + value);
}

void MULTIPLY_PACK(int value,int weight,int num)
{
    if(weight * num >= C)
    {
        COMPLETE_PACK(value,weight);
        return;
    }
    else
    {
        int k = 1;
        while(k <= num)
        {
            ZERO_ONE_PACK(k * value,k * weight);
            num -= k;
            k<<=1;
        }
        if(num)
            ZERO_ONE_PACK(num * value,num * weight);
        return;
    }
}

int main()
{
    int x,y;///x表示价值，y表示复杂度
    while(~scanf("%d%d",&n,&C))
    {
        memset(a,0,sizeof(a));
        for(int i = 0; i < n; i ++)
        {
            scanf("%s%d%d",s,&x,&y);
            a[x][y] ++;
        }
        memset(dp,0,sizeof(dp));
        for(int i = 0; i < 11; i ++)///价值
            for(int j = 0; j < 11; j ++)
                if(a[i][j])
                    MULTIPLY_PACK(i,j,a[i][j]);
        printf("%d\n",dp[C]);
    }
    return 0;
}