hdu 3732 Ahui Writes Word

 Ahui Writes Word

Time Limit: 2000/1000 MS (Java/Others)     Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 480    Accepted Submission(s): 73

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.

 

SampleInput

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

 

SampleOutput

15

 

解题思路：数据量太大，可以转换为多重背包问题，转换后再将多重背包用2进制优化成0-1背包问题。

              #include<iostream> #include<cstring> #include<cstdio> using namespace std; int p[15000]; int w[15000]; int f[10005]; int map[15][15]; int main() { int n,c,i,j,x,y,k,t,sum; char s[100]; while(scanf("%d%d",&n,&c)!=EOF) { memset(map,0,sizeof(map)); for(i=1;i<=n;i++) { scanf("%s%d%d",s,&x,&y); map[x][y]++;//记录价值和复杂度都相同的单词个数 } k=1; for(i=0;i<=10;i++) for(j=0;j<=10;j++) { if(map[i][j]>=1) { sum=1,t=1; while(sum<=map[i][j])//用1，2，4，…，2^k 将多重背包转化为0-1背包。 如13=1+2+4+6 { p[k]=t*i; w[k++]=t*j; t*=2; sum+=t; } if(sum-t<map[i][j]) { p[k]=i*(map[i][j]-sum+t); w[k++]=j*(map[i][j]-sum+t); } //printf("%d %d ",i,j); } } memset(f,0,sizeof(f)); for(i=1;i<k;i++) { for(j=c;j>=w[i];j--) { if(f[j]<f[j-w[i]]+p[i]) f[j]=f[j-w[i]]+p[i]; } } printf("%d/n",f[c]); } return 0; }