杭电3732 Ahui Writes Word （多重背包问题）

Ahui Writes Word

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 2076 Accepted Submission(s): 757



[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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)

[align=left]Output[/align]
Output the maximum value in a single line for each test case.

[align=left]Sample Input[/align]

5 20
go 5 8
think 3 7
big 7 4
read 2 6
write 3 5

[align=left]Sample Output[/align]

15
代码：这道题不能用01背包问题解决，会超时，将其化为多重背包问题 #include<stdio.h>
#include<string.h>
int max(int x,int y)
{
if(x > y)
return x;
else
return y ;
}
char a[100001] ;
int w[12][12] ;
int f[100001] ;
int va[100001] ;
int v[100001] ;
int N = 0 ,C = 0;
//完全背包
void completepack(int x ,int y)
{
for(int i = y ; i <= C ; i++ )
f[i] = max(f[i],f[i-y] + x) ;
}
//01背包
void zeroonepack(int x,int y)
{
for(int i = C ; i >= y ; i-- )
f[i] = max(f[i],f[i-y] + x) ;
}
//多重背包，这里需要用到二进制优化
void multipack(int x,int y,int n)
{
if(n * y >= C)
completepack(x,y) ;
else
{
int k = 1 ;
while(k < n)
{
zeroonepack(k * x , k * y) ;
n -= k ;
k *= 2 ;
}
zeroonepack(n * x , n * y) ;
}
}
int main()
{
while(~scanf("%d%d",&N,&C))
{
int i = 0 , j = 0 ;
int x = 0 , y = 0 ;
memset(a,0,sizeof(a));
memset(w,0,sizeof(w));
memset(f,0,sizeof(f));
memset(va,0,sizeof(va));
memset(v,0,sizeof(v));
for(i = 1 ;i <= N ; i++)
{
scanf("%s%d%d",a,&x,&y);
w[x][y]++;
}
for(i = 0 ;i < 11 ; i++)
{
for(j = 0 ; j < 11 ; j++)
{
multipack(i,j,w[i][j]);
}
}

printf("%d\n",f[C]);
}
return 0 ;
}