hdu2639 Bone Collector II

Bone Collector II

Time Limit: 2000MS Memory Limit: 32768KB 64bit IO Format: %I64d & %I64u

Submit Status

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

35 10 21 2 3 4 55 4 3 2 15 10 121 2 3 4 55 4 3 2 15 10 161 2 3 4 55 4 3 2 1 

 

Sample Output

1220 

 

Source

百万秦关终属楚

 

 

 

分析：

0-1背包，这里要求的是第K优解。也是模板题，需要进一步理解透彻。

ac代码：

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;

struct Node
{
    int price;
    int val;
} node[1005];

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n,v,k,i,dp[1005][31] = {0},a[31],b[31];
        scanf("%d%d%d",&n,&v,&k);
        for(i = 0; i<n; i++)
            scanf("%d",&node[i].price);
        for(i = 0; i<n; i++)
            scanf("%d",&node[i].val);
        int j;
        for(i = 0; i<n; i++)
        {
            for(j = v; j>=node[i].val; j--)
            {
                int cnt = 0,d;
                for(d = 1; d<=k; d++)//分别将放入第i个石头与不放第i个石头的结果存入a,b,数组之中
                {
                    a[d] = dp[j-node[i].val][d]+node[i].price;
                    b[d] = dp[j][d];
                }
                int x,y,z;
                x = y = z = 1;
                a[d] = b[d] = -1;
                while(z<=k && (x<=k || y<=k))//循环找出前K个的最优解
                {
                    if(a[x] > b[y])
                    {
                        dp[j][z] = a[x];
                        x++;
                    }
                    else
                    {
                        dp[j][z] = b[y];
                        y++;
                    }
                    if(dp[j][z]!=dp[j][z-1])
                    z++;
                }
            }
        }
        printf("%d\n",dp[v][k]);
    }

    return 0;
}