Bone Collector II（动态规划-01背包-第K优解）

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最新推荐文章于 2024-07-19 15:39:37 发布

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分类专栏： # 动态规划 ★ ACM ★

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本文介绍了一种解决01背包问题中寻找第K优解的方法，通过动态规划和额外的一维数组来记录前i个物品的第j大价值，实现了在给定数量的骨头、背包容量和需要找到的K值的情况下，求出第K大的总价值。

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Bone Collector II（动态规划-01背包-第K优解）

source：hduoj 2639
Time Limit: 5000/2000 MS (Java/Others)
Memory Limit: 32768/32768 K (Java/Others)

描述

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 $K - t h$ 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

题意

和有限容量几个物品内挑最大价值不同的是这题要求你挑出第K大值。
这时就要重新加一维数组，把 $d p [i] 改为 d p [i] [j]$ 用来表示前 $i$ 个物品的第 $j$ 大值。
然后每枚举一个物品都要维护对应的第j大值。

#include <bits/stdc++.h>
#define maxn 1005
#define inf 0x3f3f3f3f
#define _for(i, a) for(int i = 0; i < (a); i++)
#define _rep(i, a, b) for(int i = (a); i <= (b); i++)
using namespace std;
int n, V, k;
int v[maxn], w[maxn];
int dp[maxn][maxn];
int a[maxn], b[maxn];
int main() {
	int T;
	cin >> T;
	_for(q, T) {
		cin >> n >> V >> k;
		_for(i, n) cin >> v[i];
		_for(i, n) cin >> w[i];
		memset(dp, 0, sizeof(dp));
		_for(i, n) {
			for (int j = V; j >= w[i]; j--) {
				int p;
				for (p = 0; p < k; p++) {
					a[p] = dp[j - w[i]][p] + v[i];
					b[p] = dp[j][p];
				}
				a[p] = b[p] = -1;
				int x = 0, y = 0, z = 0;
				while (z <= k && (a[x] != -1 || b[y] != -1)) {
					if (a[x] > b[y]) {
						dp[j][z] = a[x++];
					}
					else {
						dp[j][z] = b[y++];
					}
					if (dp[j][z] != dp[j][z - 1]) z++;
				}
			}
		}
		cout << dp[V][k - 1] << "\n";
	}
}