題目:http://poj.org/problem?id=2299
Ultra-QuickSort
Time Limit: 7000MS | Memory Limit: 65536K |
Total Submissions: 49165 | Accepted: 17984 |
Description
In this problem, you have to analyze a particular sorting algorithm. The algorithm processes a sequence of n distinct integers by swapping two adjacent sequence elements until the sequence is sorted in ascending order. For the input sequence
9 1 0 5 4 ,
Ultra-QuickSort produces the output
0 1 4 5 9 .
Your task is to determine how many swap operations Ultra-QuickSort needs to perform in order to sort a given input sequence.
Input
The input contains several test cases. Every test case begins with a line that contains a single integer n < 500,000 -- the length of the input sequence. Each of the the following n lines contains a single integer 0 ≤ a[i] ≤ 999,999,999, the i-th input sequence element. Input is terminated by a sequence of length n = 0. This sequence must not be processed.
Output
For every input sequence, your program prints a single line containing an integer number op, the minimum number of swap operations necessary to sort the given input sequence.
Sample Input
5
9
1
5
4
3
1
2
3
Sample Output
60
本來想用這題來練習線段樹的,結果怎麼也想不到那裡去。後來看别人的代碼才明白每個葉子節點存儲的不是依據dex存儲val[i]而是依據dex在相應位置存儲1,然後根據排好序的序列由val[i]追蹤dex,左邊的1都加起來(逆序數),然後去掉dex的1,如此反複進行就能到達目的。送出代碼時我可沒這麼做,用歸并排序累加每一次相鄰的交換數就是最終的結果。(臨時在百度百科上學習了歸并排序||-_-)
冒泡排序的時間的複雜度是O(n^2).
冒泡排序算法的運作如下:(從後往前)
比較相鄰的元素。如果第一個比第二個大,就交換他們兩個。
對每一對相鄰元素作同樣的工作,從開始第一對到結尾的最後一對。在這一點,最後的元素應該會是最大的數。
針對所有的元素重複以上的步驟,除了最後一個。
持續每次對越來越少的元素重複上面的步驟,直到沒有任何一對數字需要比較。
并歸排序的時間複雜度:O(nlogn)
設有數列{6,202,100,301,38,8,1}
初始狀态:6,202,100,301,38,8,1
第一次歸并後:{6,202},{100,301},{8,38},{1},比較次數:3;
第二次歸并後:{6,100,202,301},{1,8,38},比較次數:4;
第三次歸并後:{1,6,8,38,100,202,301},比較次數:4;
總的比較次數為:3+4+4=11,;
逆序數為14;
歸并排序的優勢:分治法,已比較了的小部分不再比較,如:{8,38},{1}變成{1,8,38}隻比較1和8後{1}的長度成為0,不再比較,直接寫下{8,38}。
#include <iostream>
#include <cstdio>
using namespace std;
const int maxn=5e5+10;
int a[maxn],b[maxn];
long long sum;
void Merge(int sourceArr[],int tempArr[],int startdex,int middex,int enddex)
{
int i = startdex,j=middex+1,k = startdex;
while(i<=middex && j<=enddex)
{
if(sourceArr[i]<sourceArr[j])
tempArr[k++] = sourceArr[i++];
else
tempArr[k++] = sourceArr[j++],sum+=(middex+1-i); //if sum++; not adjacent exchange
}
while(i!=middex+1)
tempArr[k++] = sourceArr[i++];
while(j!=enddex+1)
tempArr[k++] = sourceArr[j++];
for(i=startdex;i<=enddex;i++)
sourceArr[i] = tempArr[i];
}
void MergeSort(int sourceArr[],int tempArr[],int startdex,int enddex)
{
int middex;
if(startdex<enddex)
{
middex=(startdex+enddex)/2;
MergeSort(sourceArr,tempArr,startdex,middex);
MergeSort(sourceArr,tempArr,middex+1,enddex);
Merge(sourceArr,tempArr,startdex,middex,enddex);
}
}
int main()
{
//freopen("cin.txt","r",stdin);
int n;
while(cin>>n&&n){
sum=0;
for(int i=0;i<n;i++) scanf("%d",&a[i]);
MergeSort(a,b,0,n-1);
printf("%lld\n",sum);
}
return 0;
}