hdu 3732 Ahui Writes Word 多重背包 小心超时

Ahui Writes Word

Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1052Accepted Submission(s): 399

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

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题意 ：

多重背包 多重的含义 是指同种的东西有一定的个数 即有相同的东西出现 对于相同的每个都可以取一次 可以同时取

题目意识是 输入 单词个数n 总的消耗c

再输入不同的单词 以及每个单词的价值和消耗 （这个字符串起不到任何作用 在本题中）

问能得到的最大价值

思路

本题主要卡人的地方是数据大 有可能会超时

有100000个

但是可以看到物品的价值可以和消耗最多为10 那么总共才100中物品

所以每种物品都可能有很多重复的

所以要二进制优化一下

#include<stdio.h>
#include<string.h>
struct haha
{
	int v;
	int val;
}a[100010];
int map[105][105];
int bag[10010],v;
void _01bag(int n)
{
	int i,j;
	   memset(bag,0,sizeof(bag));
        for(i=0;i<n;i++)
            for(j=v;j>=a[i].v;j--)
                if(bag[j]<bag[j-a[i].v]+a[i].val)
                    bag[j]=bag[j-a[i].v]+a[i].val;
        printf("%d\n",bag[v]);

}
int main()
{
	int n,x,y,i,j,k,temp;
	char s[20];
	while(scanf("%d %d",&n,&v)!=EOF)
	{
		memset(map,0,sizeof(map));
         for(i=1;i<=n;i++)
		 {
			 scanf("%s %d %d",s,&x,&y);//x是val y是v
			 map[x][y]++;
		 }
		 k=0;
		 for(i=0;i<=10;i++)
			 for(j=0;j<=10;j++)
			 {
                    if(map[i][j]==0) continue;
					temp=1;
					while(map[i][j]>temp)
					{
                          a[k].v=j*temp;
						  a[k++].val=i*temp;
						  map[i][j]-=temp;
						  temp=temp*2;
					}
                    a[k].v=j*map[i][j];
					a[k++].val=i*map[i][j];
			 }
			 _01bag(k);

	}

 return 0;
}