Limited Insertion

本文探讨了一种名为有限插入序列的算法挑战,该算法要求在空序列中通过N次操作,每次选择一个整数并将其插入到特定位置,最终判断是否能够得到目标序列。文章详细解释了问题背景、约束条件、输入输出格式,并提供了一个C++实现示例,通过倒序遍历和暴力判断的方法来解决这个问题。

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问题 A

***:

Limited Insertion

题目描述

Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1≤j≤i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to
b after N operations. If it is, show one possible sequence of operations that achieves it.

Constraints

·All values in input are integers.
·1≤N≤100
·1≤bi≤N

输入

Input is given from Standard Input in the following format:

N
b1 … bN

输出

If there is no sequence of N operations after which a would be equal to b, print -1. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
样例输入
3
1 2 1

样例输出

1
1
2

提示

In this sequence of operations, the sequence a changes as follows:
·After the first operation: (1)
·After the second operation: (1,1)
·After the third operation: (1,2,1)

思路

这题题目比较水,数据范围不大,就直接暴力倒叙判断即可。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=100;
struct code
{
    int num;
    int dis;
    int pos;
}a[N+5];
int b[N+5];
int c[N+5];
int main()
{
    int x;
    cin>>x;
    for(int i=1;i<=x;i++)
    {
        scanf("%d",&a[i].num);
        a[i].pos=i;
    }
    for(int i=1;i<=x;i++)
    {
        a[i].dis=0;
        /*printf("%d\n",a[i].pos-a[i].dis);*/
    }
    int n=0,flag;
    for(int i=0;i<x;i++)
    {
        flag=0;
        for(int j=x;j>0;j--)
        {
            /*cout<<a[j].pos-a[j].dis<<endl;
            cout<<a[j].num<<endl;*/
            if(a[j].num==a[j].pos-a[j].dis&&c[j]==0)
            {
                b[n++]=a[j].num;
                c[j]=1;
                flag=1;
                for(int k=j+1;k<=x;k++)
                {
                    a[k].dis++;
                }
                break;
            }
        }
        if(flag==0)
        {
            break;
        }
    }
    if(flag==0)
    {
        printf("-1\n");
    }
    else
    {
        for(int i=x-1;i>=0;i--)
        {
            printf("%d\n",b[i]);
        }
    }
    return 0;
}
### TSP (Traveling Salesman Problem) Code Implementation For the Traveling Salesman Problem, an efficient approach is using a greedy heuristic algorithm like the farthest insertion method for solving problems with more than ten points. For fewer points, brute force methods are applied due to their simplicity and effectiveness on small datasets[^1]. Below demonstrates how one might implement such algorithms. #### Farthest Insertion Heuristic Algorithm Example: This Python function implements the farthest insertion heuristic specifically designed for tackling larger instances of TSP efficiently while ensuring good performance without exhaustive search complexity. ```python import numpy as np from scipy.spatial import distance_matrix def nearest_node(unvisited, distances): min_dist = float('inf') closest_node = None for node in unvisited: dist = sum(distances[node]) if dist < min_dist: min_dist = dist closest_node = node return closest_node def tsp_far_insertion(points): n = len(points) # Calculate all pairwise Euclidean distances between cities. dists = distance_matrix(points, points) tour = [0] # Start at first city unvisited = set(range(1, n)) while unvisited: last = tour[-1] next_city = max((city for city in range(n) if city not in tour), key=lambda c: dists[last][c]) pos = min( ((i, dists[tour[i]][next_city] + dists[next_city][tour[(i+1)%n]] - dists[tour[i]][tour[(i+1)%n]]) for i in range(-len(tour), len(tour))), key=lambda x:x[1])[0] tour.insert(pos % len(tour)+1,next_city) unvisited.remove(next_city) cost = sum(dists[tour[i], tour[(i+1)%n]] for i in range(len(tour))) return tour, cost ``` The above code snippet defines two functions; `nearest_node` selects nodes based on minimum total distance criteria whereas `tsp_far_insertion` constructs tours iteratively adding furthest reachable vertices until completion. This strategy aims to minimize overall travel length effectively especially suitable for large scale scenarios beyond simple enumeration techniques feasible only within limited scope. #### Brute Force Method for Small Instances (<10 Points): When dealing with smaller sets containing less than or equal to nine elements, employing brute-force strategies becomes computationally viable offering optimal solutions albeit slower runtime compared against heuristics used previously mentioned. ```python from itertools import permutations def tsp_bruteforce(points): best_tour = None shortest_path_length = float('Inf') for perm in permutations(range(len(points))): path_length = sum(np.linalg.norm(points[perm[i]] - points[perm[i+1]]) for i in range(len(points)-1)) \ + np.linalg.norm(points[perm[-1]] - points[perm[0]]) if path_length < shortest_path_length: shortest_path_length = path_length best_tour = list(perm) return best_tour,shortest_path_length ``` In this version, every possible permutation gets evaluated calculating corresponding lengths before identifying minimal route covering each point exactly once returning both sequence alongside its associated metric value representing traveled span during traversal cycle.
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