Quick Sort
Let‘s arrange a deck of cards. Your task is to sort totally n cards. A card consists of a part of a suit (S, H, C or D) and an number. Write a program which sorts such cards based on the following pseudocode:
Partition(A, p, r) 1 x = A[r] 2 i = p-1 3 for j = p to r-1 4 do if A[j] <= x 5 then i = i+1 6 exchange A[i] and A[j] 7 exchange A[i+1] and A[r] 8 return i+1 Quicksort(A, p, r) 1 if p < r 2 then q = Partition(A, p, r) 3 run Quicksort(A, p, q-1) 4 run Quicksort(A, q+1, r)
Here, A is an array which represents a deck of cards and comparison operations are performed based on the numbers.
Your program should also report the stability of the output for the given input (instance). Here, ‘stability of the output‘ means that: cards with the same value appear in the output in the same order as they do in the input (instance).
Input
The first line contains an integer n, the number of cards.
n cards are given in the following lines. Each card is given in a line and represented by a pair of a character and an integer separated by a single space.
Output
In the first line, print the stability ("Stable" or "Not stable") of this output.
In the following lines, print the arranged cards in the same manner of that of the input.
Constraints
- 1 ≤ n ≤ 100,000
- 1 ≤ the number of a card ≤ 109
- There are no identical card in the input
Sample Input 1
6 D 3 H 2 D 1 S 3 D 2 C 1
Sample Output 1
Not stable D 1 C 1 D 2 H 2 D 3 S 3
Sample Input 2
2 S 1 H 1
Sample Output 2
Stable S 1 H 1这道题,还要看排序结果是不是稳定的,已经归并排序是稳定的,所以就拿归并排序的结果与快速排序的结果对比就好了
#include<iostream> #include<cstring> #include<stack> #include<cstdio> #include<cmath> using namespace std; #define MAX 100005 #define INF 2E9 struct Card{ char suit; int value; }; struct Card L[MAX/2+2],R[MAX/2+2]; void merge(struct Card A[],int n,int left,int mid,int right) { int i,j,k; int n1=mid-left; int n2=right-mid; for(int i=0;i<n1;i++) L[i]=A[left+i]; for(int i=0;i<n2;i++) R[i]=A[mid+i]; L[n1].value=R[n2].value=INF; i=j=0; for(k=left;k<right;k++) { if(L[i].value<=R[j].value) { A[k]=L[i++]; } else A[k]=R[j++]; } } void mergeSort(struct Card A[],int n,int left,int right) { int mid; if(left+1<right) { mid=(left+right)/2; mergeSort(A,n,left,mid); mergeSort(A,n,mid,right); merge(A,n,left,mid,right); } } int partition(struct Card A[],int n,int p,int r) { int i,j; struct Card t,x; x=A[r]; i=p-1; for(j=p;j<r;j++) { if(A[j].value<=x.value) { i++; t=A[i];A[i]=A[j];A[j]=t; } } t=A[i+1];A[i+1]=A[r];A[r]=t; return i+1; } void quickSort(struct Card A[],int n,int p,int r) { int q; if(p<r) { q=partition(A,n,p,r); quickSort(A,n,p,q-1); quickSort(A,n,q+1,r); } } int main() { int n,v; struct Card A[MAX],B[MAX]; char S[10]; int stable=1; cin>>n; for(int i=0;i<n;i++) { scanf("%s %d",S,&v); A[i].suit=B[i].suit=S[0]; A[i].value=B[i].value=v; } mergeSort(A,n,0,n); quickSort(B,n,0,n-1); for(int i=0;i<n;i++) if(A[i].suit!=B[i].suit) stable = 0; if(stable==1) cout<<"Stable"<<"\n"; else cout<<"Not stable"<<"\n"; for(int i=0;i<n;i++) { cout<<B[i].suit<<" "<<B[i].value<<"\n"; } return 0; }
原文地址:https://www.cnblogs.com/hh13579/p/10806699.html