HDU3732 - Ahui Writes Word - 动态规划之多重背包+二进制优化【经典】

1.题目描述：

Ahui Writes Word

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3080    Accepted Submission(s): 1085

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

 

Recommend

notonlysuccess

 

2.题意概述：

给出一堆单词，每个单词有val (价值)、cos (复杂度)，问消耗c的复杂度能够获得的最大价值。

3.解题思路：

就是一个多重背包问题，开始用01背包交果断超时，这时候学以致用，可以把多重背包拆分为二进制，也是一道经典的多重背包应用。

4.AC代码：

#include <stdio.h>
#include <string.h>
#include <algorithm>
#define maxn 10100
#define N 15
using namespace std;
int dp[maxn], weight[maxn], value[maxn], num[N][N];

int main()
{
	int n, c;
	while (scanf("%d%d", &n, &c) != EOF)
	{
		int val, price, cnt = 0;
		char ch[15];
		memset(dp, 0, sizeof(dp));
		memset(weight, 0, sizeof(weight));
		memset(value, 0, sizeof(value));
		memset(num, 0, sizeof(num));
		for (int i = 1; i <= n; i++)
		{
			scanf("%s%d%d", ch, &val, &price);
			num[val][price]++;
		}
		/*二进制大法好*/
		for (int i = 0; i <= 10; i++)
			for (int j = 0; j <= 10; j++)
			{
				int tmp = num[i][j];
				for (int k = 1; k <= tmp; k *= 2)
				{
					value[cnt] = i * k;
					weight[cnt++] = j * k;
					tmp -= k;
				}
				if (tmp)
				{
					value[cnt] = i * tmp;
					weight[cnt++] = j * tmp;
				}
			}
		for (int i = 0; i < cnt; i++)
			for (int j = c; j >= weight[i]; j--)
				dp[j] = max(dp[j - weight[i]] + value[i], dp[j]);  // 对于每件物品尝试拿与不拿
		printf("%d\n", dp[c]);
	}
    return 0;
}