【背包专题】F

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.

InputThe 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) 
OutputOutput 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)”

 题意：给你n个物品的价值和费用，在给定的总费用内输出最大价值。

思路：数据过大，01背包tle,由于题目给出了(0 ≤ Vi , Ci ≤ 10)这一条件，说明有大量价值和费用重复，我们可以考虑多重背包，也可以用二进制数优化后转为01背包，数量的存储可以用一个二维数组进行存储。

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define inf 0x3f3f3f3f
#define N 10010
int dp[N],w2[N],v2[N],value[N*10],weight[N*10],num[20][20];
char ch[20];
int n,v;
int main()
{
    int i,k,j,ans,x,y;
    while(scanf("%d%d",&n,&v)!=EOF)
    {
        memset(num,0,sizeof(num));
        for(i = 1; i <= n; i ++)
        {
            scanf("%s%d%d",ch,&value[i],&weight[i]);
            num[value[i]][weight[i]]++;
        }
        memset(dp,0,sizeof(dp));
        ans = 0;
        for(i = 1; i <= n; i ++)
        {
            x = value[i],y = weight[i];
            for(j = 1; num[x][y]>0; j*=2)
            {
                k = min(j,num[x][y]);
                w2[++ans] = k*y;
                v2[ans] = k*x;
                num[x][y]-= k;
            } 
        }
        for(i = 1; i <= ans; i ++)
        {
            for(j = v; j >= w2[i]; j --)
                dp[j] = max(dp[j],dp[j-w2[i]]+v2[i]);
        }
        printf("%d
",dp[v]);
    }
    return 0;
 }