POJ 2299 求逆序对（归并排序或树状数组）

Ultra-QuickSort

Time Limit: 7000MS		Memory Limit: 65536K
Total Submissions: 45290		Accepted: 16440

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

Source

Waterloo local 2005.02.05

归并排序做法：

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cstdlib>
#include<string>
using namespace std;
int num[ 500007],num1[500007],num2[5000007];
int n;
long long ans=0;
void get(int left,int mid,int right)
{
    for(int i=left;i<=right;i++)  num1[i]=num[i];
    int sum=(mid-left)+1;
    int i=left,j=mid+1,k=left;
    while(i<=mid&&j<=right){
        if(num1[i]<=num1[j]){
            num[k++]=num1[i++];
        }
        else
        {
          num[k++]=num1[j++];
          ans+=(mid-i+1);
        }
    }
    while(i<=mid){num[k++]=num1[i++];}
    while(j<=right){num[k++]=num1[j++];}
}

void merge1(int l,int r)
{
      if(l<r){

        int m=(l+r)>>1;
        merge1(l,m);
        merge1(m+1,r);
        get(l,m,r);
    }
}
int main()
{
 //  freopen("in.txt","r",stdin);
    while(1){
        scanf("%d",&n);
        ans=0;
        if(n==0) break;
        for(int i=0;i<n;i++)
            scanf("%d",&num[i]);
        merge1(0,n-1);
        printf("%I64d
",ans);
    }
    return 0;
}

树状数组做法：

#include <iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#define MAX  500010
#define LL long long
using namespace std;
int num[MAX],tree[MAX];
int n;
LL ans=0;
struct node
{
    int value,sum;
};
node arr[MAX];
int lowbit(int value)
{
    return value&(-value);
}
void build(int value)
{
  while(value<=n){
    tree[value]+=1;
    value+=lowbit(value);
  }
}
int Get_sum(int value)
{
    int result=0;
  for(int i=value;i>=1;i-=lowbit(i))
    {
        result+=tree[i];
    }
    return result;
}
bool cmp(node n1,node n2)
{
    return n1.value<n2.value;
}
int main()
{
    //freopen("in.txt","r",stdin);
 while(scanf("%d",&n)==1 && n)
    {
        ans=0;
        memset(tree,0,sizeof(tree));
        for(int i=1;i<=n;i++) scanf("%d",&arr[i].value),arr[i].sum=i;
        sort(arr+1,arr+n+1,cmp);
        for(int i=1;i<=n;i++)
        {
            num[arr[i].sum]=i;
        }
        for(int i=1;i<=n;i++)
        {
            build(num[i]);
            ans+=i-Get_sum(num[i]);
        }
        printf("%I64d
",ans);
    }
    return 0;
}