Ultra-QuickSort POJ - 2299 （逆序对）

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
0
5
4
3
1
2
3
0

Sample Output

6
0
 
题意：将一个序列从小到大排序，如果只能交换相邻的数，最少需要交换多少次
思路：和冒泡排序一样，一个数需要交换的次数就是它的逆序对数，所以就是求总的逆序对的个数
 
求逆序对可以用两种方法
①归并排序：

 #include<cstdio>
 #include<iostream>
 using namespace std;
 
 int n;
 const int maxn = 5e5+;
 int num[maxn];
 typedef long long ll;
 
 ll Mersort(int l,int r)
 {
     int mid = (l+r)/;
     int i=l,j=mid+;
     int b[r-l+];
     int k=;
     ll ans = ;
     while(i <= mid && j <= r)
     {
         if(num[i] <= num[j])
             b[k++] = num[i++];
         else
             b[k++] = num[j++],ans+=mid-i+;
     }
     while(i <= mid)
     {
         b[k++] = num[i++];
     }
     while(j <= r)
     {
         b[k++] = num[j++];
     }
     for(int i=l; i<=r; i++)
     {
         num[i] = b[i-l+];
     }
     return ans;
 }
 
 int Merge(int l,int r,ll  &ans)
 {
     int mid = (l+r)/;
     if(l == r)
         return ;
     Merge(l,mid,ans);
     Merge(mid+,r,ans);
     ans += Mersort(l,r);
 }
 int main()
 {
     while(~scanf("%d",&n) && n)
     {
         for(int i=; i<=n; i++)
             scanf("%d",&num[i]);
         ll  ans = ;
         Merge(,n,ans);
         printf("%lld\n",ans);
     }
 }

②树状数组：（要注意离散，离散可以二分，也可以map）

 #include<cstdio>
 #include<iostream>
 #include<algorithm>
 #include<string.h>
 using namespace std;
 
 const int maxn = 5e5+;
 int n;
 int tree[maxn];
 typedef long long ll;
 
 int lowbit(int x)
 {
     return x&(-x);
 }
 
 void add(int x)
 {
     for(int i=x;i<=n;i+=lowbit(i))
     {
         tree[i]++;
     }
 }
 
 int Query(int x)
 {
     int ans = ;
     for(int i=x;i>;i-=lowbit(i))
     {
         ans+=tree[i];
     }
     return ans;
 }
 int query(int x,int n,int *b)
 {
     return lower_bound(b+,b++n,x) - b;
 }
 int main()
 {
     while(~scanf("%d",&n) && n)
     {
         memset(tree,,sizeof(tree));
         int a[n+],b[n+];
         for(int i=;i<=n;i++)scanf("%d",&a[i]),b[i] = a[i];
         sort(b+,b++n);
         int len = unique(b+,b++n)-b-;
         ll ans = ;
         for(int i=;i<=n;i++)
         {
             int pos = query(a[i],len,b);
             add(pos);
             ans +=  pos -  - Query(pos-);
         }
         printf("%lld\n",ans);
     }
 }