HDU3732 背包DP

Ahui Writes Word

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2853    Accepted Submission(s): 1018

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

题意：

每背一个单词有价值和花费，给出单词数和最大花费，问得到的最大价值。

代码：

//很明显是01背包问题，但本题数据太大普通的01背包会超时，由于v和c都不大于10，最多有11*11组数据，统计每组数据有多少个，所以本题可以看做是多重背包
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int f[10004];
int N,C;
void zeroonepack(int v,int m,int ttl)
{
    for(int i=ttl;i>=v;i--)
    f[i]=max(f[i],f[i-v]+m);
}
void complitpack(int v,int m,int ttl)
{
    for(int i=v;i<=ttl;i++)
    f[i]=max(f[i],f[i-v]+m);
}
void multipack(int v,int m,int c,int ttl)
{
    if(c*v>=ttl)
    complitpack(v,m,ttl);
    else
    {
        int k=1;
        while(k<c)
        {
            zeroonepack(k*v,k*m,ttl);
            c-=k;
            k*=2;
        }
        zeroonepack(c*v,c*m,ttl);
    }
}
int main()
{
    char s[15];
    int val,cont;
    int eg[12][12];
    while(scanf("%d%d",&N,&C)!=EOF){
    memset(f,0,sizeof(f));
    memset(eg,0,sizeof(eg));
    for(int i=1;i<=N;i++)
    {
        scanf("%s%d%d",s,&val,&cont);
        eg[val][cont]++;
    }
    for(int i=0;i<=10;i++)
    {
        for(int j=0;j<=10;j++)
        {
            if(eg[i][j]==0) continue;
            multipack(j,i,eg[i][j],C);
        }
    }
    printf("%d
",f[C]);
    }
    return 0;
}