杭电 2639 Bone Collector II【01背包第k优解】

解题思路：对于01背包的状态转移方程式f[v]=max(f[v],f[v-c[i]+w[i]]);其实01背包记录了每一个装法的背包值，但是在01背包中我们通常求的是最优解，

即为取的是f[v],f[v-c[i]]+w[i]中的最大值,但是现在要求第k大的值，我们就分别用两个数组保留f[v]的前k个值，f[v-c[i]]+w[i]的前k个值，再将这两个数组合并，取第k名。

即f的数组会增加一维。

http://blog.csdn.net/lulipeng_cpp/article/details/7584981这个讲得很详细

反思：01背包没有理解，即分别用两个数组去存放f[v],f[v-c[i]]+w[i]的前k个值时，这k个值就是有序的，所以合并起来也是有序的，至于为什么是有序的，可以再看这个状态转移方程

for(i=1;i<=n;i++)

{

for(j=v;j>=c[i];j--)

f[v]=max(f[v],f[v-c[i]+w[i]]);//此时包的价值取决于上一个包有没有放进去的决策，不管那个包有没有放进去，当前状态的f[v]都是这两个值的最大值，所以

                                              从1--v,f[v]是递增的。

}

用一个简单的例子来模拟一下

有一个容量为10的包，现在有3件物品，

重量   价值

3      4

4      5

5      6

f[j]	j	1	2	3	4	5	6	7	8	9	10
i	1	0	0	4	4	4	4	4	4	4	4
	2	0	0	4	5	5	5	9	9	9	9
	3	0	0	4	5	6	6	6	6	11	11

可以看到当j的取值从1到n的时候，f[v]的值是递增的，

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

 

#include<stdio.h>
int c[1010],w[1010];
int main()
{
	int ncase,n,v,k,i,j,x,y,z,t;
	scanf("%d",&ncase);
	while(ncase--)
	{
		int f[1010][50]={0};
		int a[50],b[50];
		scanf("%d %d %d",&n,&v,&k);
		for(i=1;i<=n;i++) scanf("%d",&w[i]);
		for(i=1;i<=n;i++) scanf("%d",&c[i]);
		
		for(i=1;i<=n;i++)
		{
			for(j=v;j>=c[i];j--)
			{
				for(t=1;t<=k;t++)
				{
					a[t]=f[j-c[i]][t]+w[i];
					b[t]=f[j][t];
				}
				x=y=z=1;
		        a[t]=b[t]=-1;
	    	while(z<=k&&(x<=k||y<=k))
		         {
			    if(a[x]>b[y]) 
				   f[j][z]=a[x++];
			    else
			       f[j][z]=b[y++];
			
			     if(f[j][z]!=f[j][z-1])
			     z++;
				}
		}	
		}
		printf("%d
",f[v][k]);
	}
}