HDU 2639 Bone Collector II [動態規劃 第k大01揹包]

題幹

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

題解

/**
*求第K大的揹包，要保證是嚴格遞減的
*技巧：加一維數組保存第K大，合併的時候，發現兩個
*數組即(dp[j][1...K]和dp[j-vol[i]][1...K] + val[i])都是遞減的，
*所以O(n)時間內就可以完成dp[j][1...K]數組的更新
*轉移方程：dp[j][t]表示體積爲j時，第k大的揹包的最優值(1<=t<=K)
*dp[j][1...K] = max(dp[j][1...K], dp[j-vol[i]][1...K] + val[i])(其中i表示第i個物品)
*/
#include<cstdio>
#include<cstring>

using namespace std;

const int inf = 0x3f3f3f3f;
int dp[1010][70];
int main(){
    //freopen("in.txt","r",stdin);
    int T,N,V,K;
    int tmp[70];
    int val[110];
    int vol[110];
    scanf("%d",&T);
    while(T--){
        scanf("%d%d%d",&N,&V,&K);
        for(int i=1;i<=N;i++)scanf("%d",&val[i]);
        for(int i=1;i<=N;i++)scanf("%d",&vol[i]);
        memset(dp, 0, sizeof(dp));
        for(int i=2;i<70;i++)dp[0][i] = -inf;//賦值爲-inf，表示該位置保存的數據無效
        for(int j=0;j<=V;j++)for(int i=K+1;i<70;i++)dp[j][i] = -inf;//賦值爲-inf，表示該位置保存的數據無效，
        //以此保證合併兩個數組時更方便
        for(int i=1; i<=N; i++){
            for(int j=V; j>=vol[i]; j--){
               int l = 1, r = 1, t = 1;
               //初始化tmp數組，初始爲-inf，防止因爲重複的價值使得合併時不能給後面的位置賦值
               memset(tmp,-inf,sizeof(tmp));
               //將值合併到tmp中
               while(t <= K){
                  if(dp[j][l] > dp[j-vol[i]][r] + val[i]){
                     tmp[t] = dp[j][l++];
                  }else{
                     tmp[t] = dp[j-vol[i]][r++] + val[i];
                  }
                  if(tmp[t]!=tmp[t-1])t++;
                  if(l > K && r > K)break;
               }
               //將tmp的數據更新到dp數組
               for(int i=1;i<=K;i++)dp[j][i] = tmp[i];
            }
        }
        if(dp[V][K]<0)printf("0\n");
        else printf("%d\n",dp[V][K]);
    }
    return 0;
}