hdu-2639-01揹包變形之求第k優解-Bone Collector II

Bone Collector II

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

題意：T組數據，N個骨頭，揹包容量爲V，求第K優解

第二行爲價值，第三行爲體積

分析：dp[i][j]表示體積爲i時第j優解

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int V,K;
int a[35],b[35],dp[1005][35];
int c[105],w[105],d[70];
int cmp(int a1,int b1)
{
    return a1>b1;
}
void solve(int c1,int w1)
{
    int i,l,j;
    for(i=V;i>=c1;i--)
    {
        for(j=1;j<=K;j++)
        {
            a[j]=dp[i][j];      //a[]用來存不放當前物品各個體積狀態下的價值
            b[j]=dp[i-c1][j]+w1;    //b[]用來存放當前物品各個體積狀態下的價值
            d[j-1]=a[j];
        }
        //選取a和b數組當中的前k大，用來更新dp數組，表示體積爲i時的第j優解
        for(j=1;j<=K;j++)
        {
            d[j+K-1]=b[j];
        }
        sort(d,d+2*K,cmp);
        int temp=d[0];
        dp[i][1]=temp;
        l=2;
        for(j=1;j<2*K;j++)
        {
            if(d[j]!=temp)
            {
               dp[i][l++]=d[j];
               temp=d[j];
            }
            if(l==K+1)
                break;
        }
    }
}
int main()
{
    int T,n,i,j;
    scanf("%d",&T);
    while(T--)
    {
        memset(d,0,sizeof(d));
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        memset(dp,0,sizeof(dp));
        scanf("%d%d%d",&n,&V,&K);
        for(i=0;i<n;i++)
            scanf("%d",&w[i]);
        for(i=0;i<n;i++)
            scanf("%d",&c[i]);
        for(i=0;i<n;i++)
        {
            solve(c[i],w[i]);
        }
        printf("%d\n",dp[V][K]);
    }
    return 0;
}