POJ 2299 逆序对

Crossings

Time Limit: 2 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/gym/100463

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
0
5
4
3
1
2
3
0

Sample Output

6
0

HINT

题意

逆序对求解

题解：

树状数组水

代码

 #include <cstdio>
 #include <cmath>
 #include <cstring>
 #include <ctime>
 #include <iostream>
 #include <algorithm>
 #include <set>
 #include <vector>
 #include <queue>
 #include <map>
 #include <stack>
 #define MOD 1000000007
 #define maxn 32001
 using namespace std;
 typedef __int64 ll;
 inline ll read()
 {
     ll x=,f=;
     char ch=getchar();
     while(ch<''||ch>'')
     {
         if(ch=='-')f=-;
         ch=getchar();
     }
     while(ch>=''&&ch<='')
     {
         x=x*+ch-'';
         ch=getchar();
     }
     return x*f;
 }
 //*******************************************************************

 struct ss
 {
     int v,index;
 } in[];
 int c[];
 int a[];
 int n;
 bool cmp(ss s1,ss s2)
 {
     return s1.v<s2.v;
 }
 int lowbit(int x)
 {
     return x&(-x);
 }
 int getsum(int x)
 {
     int sum=;
     while(x>)
     {
         sum+=c[x];
         x-=lowbit(x);
     }
     return sum;
 }
 void update(int x,int value)
 {
     while(x<=n)
     {
         c[x]+=value;
         x+=lowbit(x);
     }
 }
 int main()
 {

     while(scanf("%d",&n)!=EOF)
     {
         if(n==) break;
         memset(c,,sizeof(c));
         for(int i=; i<=n; i++)
         {
             in[i].v=read();
             in[i].index=i;
         }
         sort(in+,in+n+,cmp);
         for(int i=; i<=n; i++)a[in[i].index]=i; //离散化处理
         ll ans=;
         for(int i=; i<=n; i++)
         {
             update(a[i],);
             ans+=i-getsum(a[i]);
         }
         cout<<ans<<endl;
     }

     return ;
 }