【HDU-2639】Bone Collector II（背包）

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 2 31).

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

题目大意

Bone Collector的升级版，求第K大价值。

思路

基本思路不变，还是01背包问题，但是不是求背包的最大价值而是求第K大价值，可以用set，队列来维护第K大值，也可以简单的用两个数组d1,d2来记录每一个值，通过比较d1，d2的值更新dp数组，这里dp数组要开二维（注意这不是二维费用的背包问题，不要混淆）即dp[maxn][31]，第二维用来表示第几大的价值。
Bone Collector的状态转移方程是：

dp[j]=max(dp[j],dp[j-w[i]]+v[i])；

但是这题不是，而是改用一个从1~k的循环来求每一个费用下面放和不放第i的骨头的价值，即：

for(l=1;l<=k;l++){   d1[l]=dp[j][l];//不放的情况   d2[l]=dp[j-w[i]][l]+v[i];//放的情况}

接下来只要根据大小合并d1,d2就行了。这样dp[wei][k]就是背包长度为wei下第K大的价值了

代码

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn=100000+5;int dp[maxn][31],v[maxn],n,k,w[maxn],wei;int d1[maxn],d2[maxn];int main(){    int t;    scanf("%d",&t);    while(t--)    {        memset(dp,0,sizeof(dp));        scanf("%d %d %d",&n,&wei,&k);        for(int i=1;i<=n;i++)        {            scanf("%d",&v[i]);        }        for(int i=1;i<=n;i++)        {            scanf("%d",&w[i]);        }        for(int i=1;i<=n;i++)        {            for(int j=wei;j>=w[i];j--)            {                int l;                for(l=1;l<=k;l++)                {                    d1[l]=dp[j][l];                    d2[l]=dp[j-w[i]][l]+v[i];                }                d1[l]=d2[l]=-1;                int x,y,z;                x=y=z=1;                while(z<=k&&(d1[x]!=-1||d2[y]!=-1))                {                    if(d1[x]>d2[y])                    {                        dp[j][z]=d1[x++];                    }                    else                     {                        dp[j][z]=d2[y++];                    }                    if(dp[j][z-1]!=dp[j][z]) z++;                }            }        }        printf("%d\n",dp[wei][k]);    }    return 0;}