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=0,f=1;
    char ch=getchar();
    while(ch<'0'||ch>'9')
    {
        if(ch=='-')f=-1;
        ch=getchar();
    }
    while(ch>='0'&&ch<='9')
    {
        x=x*10+ch-'0';
        ch=getchar();
    }
    return x*f;
}
//*******************************************************************

struct ss
{
    int v,index;
} in[500005];
int c[500005];
int a[500005];
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=0;
    while(x>0)
    {
        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==0) break;
        memset(c,0,sizeof(c));
        for(int i=1; i<=n; i++)
        {
            in[i].v=read();
            in[i].index=i;
        }
        sort(in+1,in+n+1,cmp);
        for(int i=1; i<=n; i++)a[in[i].index]=i; //离散化处理
        ll ans=0;
        for(int i=1; i<=n; i++)
        {
            update(a[i],1);
            ans+=i-getsum(a[i]);
        }
        cout<<ans<<endl;
    }

    return 0;
}