POJ 2299 Ultra-QuickSort 题解

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Ultra-QuickSort

Time Limit: 7000MS Memory Limit: 65536KTotal Submissions: 53630 Accepted: 19693

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

59105431230

Sample Output

Source

Waterloo local 2005.02.05

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好题。归并排序求逆序对。

 1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cmath> 5 #include<algorithm> 6 #include<queue> 7 #include<cstdlib> 8 #include<iomanip> 9 #include<cassert>10 #include<climits>11 #define maxn 1000112 #define F(i,j,k) for(int i=j;i<=k;i++)13 #define FF(i,j,k) for(int i=j;i>=k;i--)14 #define inf 0x7fffffff15 #define NN 50000416 #define NL 100017 #define mem(a) memset(a, 0, sizeof(a))18 using namespace std;19 int N, A[500010], T[500010];20 __int64 ans;21 int read(){22     int x=0,f=1;char ch=getchar();23     while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}24     while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}25     return x*f;26 }27 __int64 res;28 int b[NN];29 void copy(int a[], int l, int r){30     int i;31     for (i = l; i <= r; i++){32         a[i] = b[i];33     }34 }35 void merge(int a[], int l, int mid, int r){36     int i = l;37     int j = mid + 1;38     int k = l;39     while(i <= mid && j <= r){40         if(a[i] < a[j]){41             b[k++] = a[i];42             i++;43         }else{44             b[k++] = a[j];45             j++;46             res += mid - i + 1;47         }48     }49     while(i <= mid){50         b[k++] = a[i];51         i++;52     }53     while(j <= r){54         b[k++] = a[j];55         j++;56     }57 }58 void mergeSort(int a[], int l, int r){59     if(l < r){60         int mid = (l + r) >> 1;61         mergeSort(a, l, mid);62         mergeSort(a, mid + 1, r);63         merge(a, l, mid, r);64         copy(a, l, r);65     }66 }67 int main() {68     int n, i;69     int f[NN];70     while(cin>>n&&n){71         if(n == 0) break;72         for (i = 1; i <= n; i++){73             cin>>f[i];74         }75         res = 0;76         mergeSort(f, 1, n);77         cout<<res<<endl;78     }79     return 0;80 }

poj 2299

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