HDOJ 题目2639Bone Collector II（01背包第k优解）

Bone Collector II

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 2284    Accepted Submission(s): 1196


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

 

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d[i][j][k], 表示取前i个物品，用j的费用，第k大价值是多少

在递推d[i][j][1...k]时，先获取上一个状态d[i-1][j][1...k]递推出来所有的值:

即集合A={dp[i-1][j][p]+w[i], 1<=p<=k}， 还有原来的值集合B={dp[i-1][j][p], 1<=p<=k}

然后把集合A和B中的前k大的值按从大到小顺序赋值给d[i][j][1...k]
ac代码

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
int a[110],b[110],dp[1100][1100],w[110],v[110];
int cmp(const void *a,const void *b)
{
return *(int *)b-*(int *)a;
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int n,m,r,i,j,k,na,nb,x,y,temp;
scanf("%d%d%d",&n,&m,&r);
memset(a,0,sizeof(a));
memset(b,0,sizeof(b));
memset(dp,0,sizeof(dp));
for(i=0;i<n;i++)
{
scanf("%d",&v[i]);
}
for(i=0;i<n;i++)
scanf("%d",&w[i]);
for(i=0;i<n;i++)
{
for(j=m;j>=w[i];j--)
{
na=0,nb=0;
for(k=1;k<=r;k++)
{
//if(dp[j][k])
b[nb++]=dp[j][k];
//else
a[na++]=dp[j-w[i]][k]+v[i];
}
qsort(a,na,sizeof(a[0]),cmp);//a,b从大到小排
qsort(b,nb,sizeof(b[0]),cmp);
x=0,y=0,k=1;
while(k<=r&&(x<na||y<nb))//合并排序
{
temp=0;
if(y>=nb||(x<na&&a[x]>b[y]))
{
temp=a[x++];
}
else
temp=b[y++];
if(j==1||temp!=dp[j][k-1])
{
dp[j][k]=temp;
k++;
}
}
}
}
printf("%d\n",dp[m][r]);
}
}