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01背包 HDU 2639，第K優解

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01背包 HDU 2639，第K優解

只要每次都保存前k優解就可以了

注意當價值一樣時，只算一種，所以要進行去重復的操作

用到了unique，

1 2 2 4 0 0

unique之後1 2 4 0 0 2

Bone Collector II

Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2433 Accepted Submission(s): 1277

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

Sample Output

12
2
0

Author teddy
Source 百萬秦關終屬楚

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
int dp[1111][33];
int cost[111];
int val[111];
int mark[111];

bool cmp(int a,int b){
    return a>b;
}

int main(){
    int T,n,v,i,j,k,kk;
    int con;

    while(~scanf("%d",&T)){
        while(T--){
            memset(dp,0,sizeof(dp));
            memset(cost,0,sizeof(cost));
            memset(val,0,sizeof(val));
            scanf("%d%d%d",&n,&v,&k);
            for(i=1;i<=n;i++){
                scanf("%d",&val[i]);
            }
            for(i=1;i<=n;i++){
                scanf("%d",&cost[i]);
            }
            for(i=1;i<=n;i++){
                for(j=v;j>=cost[i];j--){
                    con = 1;
                    memset(mark,0,sizeof(mark));
                    for(kk=1;kk<=k;kk++){
                        mark[con] = dp[j][kk];
                        con++;
                        mark[con] = dp[j-cost[i]][kk]+val[i];
                        con++;
                    }
                    sort(mark+1,mark+con+1,cmp);
                    unique(mark,mark+con+1);
                    for(int t=1;;t++){
                        if(mark[t] == 0){
                            for(int q=t+1;q<=66;q++){
                                mark[q] = 0;
                            }
                            break;
                        }
                    }
                    for(kk=1;kk<=k;kk++){
                        dp[j][kk] = mark[kk];
                    }
                }
            }
            printf("%d\n",dp[v][k]);
        }
    }

    return 0;
}