Greedy descent with random restarts will find the optimal solution with probability 1, given sufficiently many restarts are performed.

是的，Greedy descent with random restarts算法可以在足够多的重启次数的情况下以概率1找到最优解。该算法的基本思想是通过贪心策略来搜索解空间，在搜索过程中遇到局部最优解时，以一定概率随机重启搜索，以期望找到更优的解。随着重启次数的增加，算法逐渐接近最优解。虽然该算法可以保证在理论上找到最优解，但是实际上需要大量的计算资源和时间来完成搜索，因此在实际应用中需要权衡计算成本和解的质量。

相关问题

Variable Neighbourhood Descent (VND) Use k successor relations, ordered according to increasing size of the respective neighbourhoods N1, N2, . . . , Nk . Start with the first neighbourhood N1 and perform greedy descent until local optimum. Whenever no further improving step is found for a neighbourhood Ni where i + 1 ≤ k, continue the search in neighbourhood Ni+1. If an improvement is obtained in Ni , search goes back to N1.请用中文解释

变邻域下降（VND）是一种启发式搜索算法。该算法使用 k 个后继关系，这些关系按照相应邻域 N1、N2、…、Nk 的大小递增排序。首先在第一个邻域 N1 中进行贪心下降，直到达到一个局部最优解。每当在第 i 个邻域（其中 i+1 ≤ k）中找不到更进一步改善的步骤时，就继续在邻域 Ni+1 中搜索。如果在 Ni 中获得了改进，搜索将回到 N1。

简单地说，VND算法先在最小邻域中进行贪心下降，找到局部最优解。然后，如果在当前邻域中找不到更好的解，就转移到下一个更大的邻域中搜索，直到找到一个更好的解。如果在某个邻域中找到了更好的解，就重新从最小邻域开始搜索。这样，算法可以在多个邻域中搜索，从而更有可能找到全局最优解。

There are � n applicants and � m free apartments. Your task is to distribute the apartments so that as many applicants as possible will get an apartment. Each applicant has a desired apartment size, and they will accept any apartment whose size is clos

This problem can be solved using a greedy approach. First, sort the applicants' desired apartment sizes and the available apartment sizes in non-decreasing order. Then, iterate through the applicants and for each applicant, find the smallest available apartment that is either the same size or larger than the applicant's desired size. If such an apartment exists, assign it to the applicant and remove it from the list of available apartments. If not, move on to the next applicant. Repeat this process until all applicants have been assigned an apartment or there are no more available apartments.

Here's some sample code in Python:

n, m, k = map(int, input().split())
applicants = sorted(list(map(int, input().split())))
apartments = sorted(list(map(int, input().split())))

i = j = count = 0
while i < n and j < m:
    if abs(applicants[i] - apartments[j]) <= k:
        count += 1
        i += 1
        j += 1
    elif applicants[i] < apartments[j]:
        i += 1
    else:
        j += 1

print(count)

This code reads in the input and sorts the applicants' desired sizes and the available apartment sizes. Then, it iterates through the applicants and available apartments using two pointers, i and j respectively. If an available apartment is found that is close enough in size to an applicant's desired size, the applicant is assigned the apartment and the count is incremented. If not, the pointers are moved forward. Finally, the count of assigned apartments is printed.