1098 Insertion or Heap Sort——PAT甲级真题
1098 Insertion or Heap Sort
According to Wikipedia: Insertion sort iterates, consuming one input element each repetition, and growing a sorted output list. Each iteration, insertion sort removes one element from the input data, finds the location it belongs within the sorted list, and inserts it there. It repeats until no input elements remain.
Heap sort divides its input into a sorted and an unsorted region, and it iteratively shrinks the unsorted region by extracting the largest element and moving that to the sorted region. it involves the use of a heap data structure rather than a linear-time search to find the maximum.
Now given the initial sequence of integers, together with a sequence which is a result of several iterations of some sorting method, can you tell which sorting method we are using?
Input Specification:
Each input file contains one test case. For each case, the first line gives a positive integer N (<=100). Then in the next line, N integers are given as the initial sequence. The last line contains the partially sorted sequence of the N numbers. It is assumed that the target sequence is always ascending. All the numbers in a line are separated by a space.
Output Specification:
For each test case, print in the first line either “Insertion Sort” or “Heap Sort” to indicate the method used to obtain the partial result. Then run this method for one more iteration and output in the second line the resuling sequence. It is guaranteed that the answer is unique for each test case. All the numbers in a line must be separated by a space, and there must be no extra space at the end of the line.
Sample Input 1:
10
3 1 2 8 7 5 9 4 6 0
1 2 3 7 8 5 9 4 6 0
Sample Output 1:
Insertion Sort
1 2 3 5 7 8 9 4 6 0
Sample Input 2:
10
3 1 2 8 7 5 9 4 6 0
6 4 5 1 0 3 2 7 8 9
Sample Output 2:
Heap Sort
5 4 3 1 0 2 6 7 8 9
题目大意:给你一组初始序列和中间序列,让你根据序列来判断是插入排序还是堆排序。然后输出中间序列的下一趟排序
解法1:按照题目所给要求对插入排序的每一步进行模拟,每一趟的排序结果和中间序列进行比较,如果相同则为插入排序,否则为堆排序。注意:因为序列是按照从小到大排列,所以根据堆的性质我们要构建的堆是大根堆。
代码:
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
int init_array[N], final_array[N], tmp_array[N];
int n;
bool same_array() {
for (int i = 1; i <= n; i++) {
if (init_array[i] != final_array[i]) return false;
}
return true;
}
void print_array() {
for (int i = 1; i <= n; i++) {
cout << init_array[i];
if (i != n)
cout << " ";
else
cout << endl;
}
}
bool insert_sort() {
int i, j;
bool flag = false;
for (i = 2; i <= n; i++) {
if (i != 2 && same_array()) flag = true;
if (init_array[i] < init_array[i - 1]) {
init_array[0] = init_array[i];
for (j = i - 1; init_array[0] < init_array[j]; j--)
init_array[j + 1] = init_array[j];
init_array[j + 1] = init_array[0];
if (flag) {
return true;
}
}
}
return false;
}
void down(int u, int sizes) {
int t = u;
if (u * 2 <= sizes && init_array[u * 2] > init_array[t]) t = u * 2;
if (u * 2 + 1 <= sizes && init_array[u * 2 + 1] > init_array[t])
t = u * 2 + 1;
if (t != u) {
swap(init_array[t], init_array[u]);
down(t, sizes);
}
}
void up(int u) {
while (u / 2 && init_array[u / 2] > init_array[u]) {
swap(init_array[u / 2], init_array[u]);
u /= 2;
}
}
void heap_sort() {
//初建堆,因为是从小打到输出所以构建的是大顶堆
for (int i = n / 2; i >= 1; i--) down(i, n);
bool flag = false;
for (int i = n; i > 1; i--) {
if (i != n && same_array()) flag = true;
swap(init_array[1], init_array[i]);
down(1, i - 1);
if (flag) {
print_array();
return;
}
}
}
int main() {
scanf("%d", &n);
// sizes = n;
for (int i = 1; i <= n; i++) {
scanf("%d", &init_array[i]);
tmp_array[i] = init_array[i];
}
for (int i = 1; i <= n; i++) scanf("%d", &final_array[i]);
if (insert_sort()) {
puts("Insertion Sort");
print_array();
} else {
puts("Heap Sort");
for (int i = 1; i <= n; i++) init_array[i] = tmp_array[i];
heap_sort();
}
return 0;
}
解法2:参考柳神的解法 ,堆排序大致思路相同,关键的是对插入排序的优化。根据插入排序的性质可以知道结点L(i)前L(1 ~i - 1)个元素都是有序序列,结点L(i + 1, n)个元素都是无序序列。所以我们只需要在中间序列中找到结点L(i)然后比较中间序列的L(i + 1, n)和原序列的L(i + 1, n),如果相同则为插入排序,如果不同则为堆排序。如果是插入排序直接对L(1 ~ i + 1)个元素进行sort即可。
利用插入排序优化的代码如下:
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
int init_array[N], final_array[N];
int n;
void down(int h[], int u, int sizes) {
int t = u;
if (u * 2 <= sizes && h[u * 2] > h[t]) t = u * 2;
if (u * 2 + 1 <= sizes && h[u * 2 + 1] > h[t])
t = u * 2 + 1;
if (t != u) {
swap(h[t], h[u]);
down(h, t, sizes);
}
}
int main() {
scanf("%d", &n);
// sizes = n;
for (int i = 1; i <= n; i++) {
scanf("%d", &init_array[i]);
}
for (int i = 1; i <= n; i++) scanf("%d", &final_array[i]);
int p = 1;
while(p < n && final_array[p] <= final_array[p + 1]) p++;
bool flag = true;
for (int i = p + 1; i <= n; i++) {
if (init_array[i] != final_array[i]) {
flag = false;
break;
}
}
if (flag) {
puts("Insertion Sort");
sort(final_array + 1, final_array + p + 2);
}
else {
puts("Heap Sort");
int index = n;
while(index > 2 && final_array[index] > final_array[1]) index--;
swap(final_array[1], final_array[index]);
down(final_array,1, index - 1);
}
for (int i = 1; i <= n; i++) {
cout << final_array[i];
if (i != n)
cout << " ";
else
cout << endl;
}
return 0;
}
1098 Insertion or Heap Sort——PAT甲级真题的更多相关文章
- PAT甲级1098. Insertion or Heap Sort
PAT甲级1098. Insertion or Heap Sort 题意: 根据维基百科: 插入排序迭代,消耗一个输入元素每次重复,并增加排序的输出列表.在每次迭代中,插入排序从输入数据中删除一个元素 ...
- PAT甲级——1098 Insertion or Heap Sort (插入排序、堆排序)
本文同步发布在CSDN:https://blog.csdn.net/weixin_44385565/article/details/90941941 1098 Insertion or Heap So ...
- pat 甲级 1098. Insertion or Heap Sort (25)
1098. Insertion or Heap Sort (25) 时间限制 100 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yu ...
- 1098 Insertion or Heap Sort
1098 Insertion or Heap Sort (25 分) According to Wikipedia: Insertion sort iterates, consuming one in ...
- PAT 甲级真题题解(63-120)
2019/4/3 1063 Set Similarity n个序列分别先放进集合里去重.在询问的时候,遍历A集合中每个数,判断下该数在B集合中是否存在,统计存在个数(分子),分母就是两个集合大小减去分 ...
- 【PAT甲级】1098 Insertion or Heap Sort (25 分)
题意: 输入一个正整数N(<=100),接着输入两行N个数,表示原数组和经过一定次数排序后的数组.判断是经过插入排序还是堆排序并输出再次经过该排序后的数组(数据保证答案唯一). AAAAAcce ...
- PAT (Advanced Level) Practise - 1098. Insertion or Heap Sort (25)
http://www.patest.cn/contests/pat-a-practise/1098 According to Wikipedia: Insertion sort iterates, c ...
- PAT (Advanced Level) 1098. Insertion or Heap Sort (25)
简单题.判断一下是插排还是堆排. #include<cstdio> #include<cstring> #include<cmath> #include<ve ...
- PAT甲题题解1098. Insertion or Heap Sort (25)-(插入排序和堆排序)
题目就是给两个序列,第一个是排序前的,第二个是排序中的,判断它是采用插入排序还是堆排序,并且输出下一次操作后的序列. 插入排序的特点就是,前面是从小到大排列的,后面就与原序列相同. 堆排序的特点就是, ...
随机推荐
- jQuery操作CheckBox的方法(选中,取消,取值)详解
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN"> <HTML> <HEAD ...
- H3C交换机端口聚合配置
1.接入交换机: interface Ten-GigabitEthernet1/0/21 port link-mode bridge port link-type trunk port trunk p ...
- Think in Java 第四 五 章
Think in Java 第四章 控制执行流程 测试while public class whileTest { static boolean condition(){ boolean result ...
- Shiro权限项目
目录 环境配置 spring容器 springmvc freemarker mybatis shiro 工具类 TokenManager.java Result.java 功能实现 登录 注册 个人中 ...
- 机器学习算法之Kmeans算法(K均值算法)
Kmeans算法(K均值算法) KMeans算法是典型的基于距离的聚类算法,采用距离作为相似性的评价指标,即认为两个对象的距离越近,其相似度就越大.该算法认为簇是由距离靠近的对象组成的,因此把得到紧凑 ...
- mysql查询太慢,我们如何进行性能优化?
老刘是即将找工作的研究生,自学大数据开发,一路走来,感慨颇深,网上大数据的资料良莠不齐,于是想写一份详细的大数据开发指南.这份指南把大数据的[基础知识][框架分析][源码理解]都用自己的话描述出来,让 ...
- [ICPC 2018 宁夏邀请赛] A-Maximum Element In A Stack(思维)
>传送门< 前言 辣鸡网络赛,虽然我是个菜鸡,然而好几个队伍十几分钟就AK???我心态那会彻底崩了,后来群里炸了,话题直接上知乎热搜,都是2018ICPC宁夏网络赛原题,这怎么玩,拼手速? ...
- CF 1288 E. Messenger Simulator
CF 1288 E. Messenger Simulator 题目传送门 官方题解 题意想必大家都明白了这里就不赘述了,这里只想重点记录一下几种实现方法 分析 设向前移动的序列为\(a\)序列 对于没 ...
- 【bzoj 3333】排队计划(线段树)
n个数,求一次逆序对.接着有m次修改操作,把每次输入的位置p的数之后<=它的数取出来,从小到大排序后再放回空位里,求逆序对.(N,M<=500,000 , Ai<=10^9)思路:1 ...
- fzu2178礼物分配 (状压+二分)
Problem Description 在双胞胎兄弟Eric与R.W的生日会上,他们共收到了N个礼物,生日过后他们决定分配这N个礼物(numv+numw=N).对于每个礼物他们俩有着各自心中的价值vi ...