Bone Collector II 背包，01背包

Bone Collector II

Time Limit : 5000/2000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 19   Accepted Submission(s) : 17

Problem Description

The title of this problem is familiar,isn't it?yeah,if you had took part in the "Rookie Cup" competition,you must have seem this title.If you haven't seen it before,it doesn't matter,I will give you a link:

Here is the link:http://acm.hdu.edu.cn/showproblem.php?pid=2602

Today we are not desiring the maximum value of bones,but the K-th maximum value of the bones.NOTICE that,we considerate two ways that get the same value of bones are the same.That means,it will be a strictly decreasing sequence from the 1st maximum , 2nd maximum .. to the K-th maximum.

If the total number of different values is less than K,just ouput 0.

 

Input

The first line contain a integer T , the number of cases.
Followed by T cases , each case three lines , the first line contain two integer N , V, K(N <= 100 , V <= 1000 , K <= 30)representing the number of bones and the volume of his bag and the K we need. And the second line contain N integers representing the value of each bone. The third line contain N integers representing the volume of each bone.

 

Output

One integer per line representing the K-th maximum of the total value (this number will be less than 2³¹).

 

Sample Input

3
5 10 2
1 2 3 4 5
5 4 3 2 1
5 10 12
1 2 3 4 5
5 4 3 2 1
5 10 16
1 2 3 4 5
5 4 3 2 1

 

Sample Output

12
2
0

 

Author

teddy

 

Source

百万秦关终属楚

 

 

求第k大的值。要多开一维的数组。保存前k大的数。看别人的思路，自己不是很会表达，这里贴张图。记录一下，怕以后忘了。

 

再贴上自己的代码。

#include<stdio.h>
long c[1001][31],x,t1[31],t2[31];
void unite(long i)
{
long j,k,l,flag;
k=1;l=1;
for(j=1;j<=x;j++)
{
if(k>x&&l>x)
break;
while(flag==0)
{
if(t1[k]>t2[l])
{
if(t1[k]!=c[i][j-1])
{
c[i][j]=t1[k];
k++;
flag=1;
}
else
k++;
}
else
{
if(t2[l]!=c[i][j-1])
{
c[i][j]=t2[l];
l++;
flag=1;
}
else
l++;
}
if(k>x&&l>x)
break;
}
flag=0;
}
}
int main()
{
long t,n,v,i,j,k,l,a[101],b[101];
while(scanf("%ld",&t)!=EOF)
{
for(i=1;i<=t;i++)
{
scanf("%ld%ld%ld",&n,&v,&x);
for(j=1;j<=n;j++)
scanf("%ld",&a[j]);
for(j=1;j<=n;j++)
scanf("%ld",&b[j]);
for(j=0;j<=v;j++)
for(k=0;k<=x;k++)
c[j][k]=0;
t1[x+1]=-1;t2[x+1]=-1;
for(j=1;j<=n;j++)
for(k=v;k>=b[j];k--)
{
for(l=1;l<=x;l++)
{
t1[l]=c[k-b[j]][l]+a[j];
t2[l]=c[k][l];
}
unite(k);
}
printf("%ld ",c[v][x]);
}
}
}

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