POJ 2299 Ultra-QuickSort 求逆序对数（归并排序，树状数组）

Ultra-QuickSort

Time Limit: 7000MS		Memory Limit: 65536K
Total Submissions: 36281		Accepted: 13078

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

 

【题目大意】：

          求对于一数列冒泡的交换次数

【分析】：

         转化题目模型：求序列的逆序对数

      逆序对数：设A[1..n]是一个包含N个非负整数的数组。如果在i<j的情况下，有A[i]>A[j]，则(i,j)就称为A中的一个逆序对。例如，数组（3，1，4，5，2）的“逆序对”有<3,1>,<3,2><4,2><5,2>，共4个。定义：对于一个给定的数列,如果有i<j,且Ai>Aj,则称(i,j)为一逆序对. 要解决的问题是,给出一个数列,求出这个数列包含多少个逆序对

            暂时了解到的有三种方法：归并排序，树状数组和权值线段树，代码给出前两种方法

【代码1，归并排序】：

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
#include<iostream>
#include<vector>
#include<stack>
#include<queue>
using namespace std;
#define MAXN 500001
int N,a[MAXN],t[MAXN];
long long ans=0;
void merge(int left,int mid,int right)
{
      int now=0,l=left,r=mid+1;
      while(l<=mid && r<=right)
      {
            if(a[l]>a[r])
            {
                  t[now++]=a[r++];
                  ans+=mid-l+1;
            }
            else  t[now++]=a[l++];
      }
      while(l<=mid)   t[now++]=a[l++];
      while(r<=right) t[now++]=a[r++];
      for(int i=0;i<now;i++)
            a[left+i]=t[i];
}
void mergesort(int left,int right)
{
      if(left<right)
      {
            int middle=(left+right)/2;
            mergesort(left,middle);
            mergesort(middle+1,right);
            merge(left,middle,right);
      }
}
int main()
{
      //freopen("input.in","r",stdin);
	  //freopen("output.out","w",stdout); 
	  while(scanf("%d",&N)!=EOF && N!=0)
	  {
            ans=0;
            for(int i=1;i<=N;i++)
                  scanf("%d",&a[i]);
            mergesort(1,N);
            /*for(int i=1;i<=N;i++)
                  printf("%d ",a[i]);
            printf("\n");*/
            printf("%I64d\n",ans);
      }
	  //system("pause");
      return 0;
}

 

【代码2，树状数组】：

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
#include<iostream>
#include<vector>
#include<stack>
#include<queue>
using namespace std;
#define MAXN 500001
struct DATA{int old,num,v;};
DATA a[MAXN];
int N,kind=0,old[MAXN];
long long S[MAXN],ans=0;
bool cmp(DATA X,DATA Y){return X.old<Y.old;}
bool cmp1(DATA X,DATA Y){return X.num<Y.num;}
int lowbit(int x){return x&(-x);}
void add(int x,int value)
{
      int now=x;
      while(now<=kind)
      {
            S[now]+=value;
            now+=lowbit(now);
      }
}
long long getsum(int x)
{
      int now=x;
      long long tmp=0;
      while(now>0)
      {
            tmp+=S[now];
            now-=lowbit(now);
      }
      return tmp;
}
int main()
{
      //freopen("input.in","r",stdin);
	  //freopen("output.out","w",stdout); 
	  while(scanf("%d",&N)!=EOF && N!=0)
      {
            memset(S,0,sizeof(S));
            ans=0;
            kind=0;
            for(int i=1;i<=N;i++)
            {
                  scanf("%d",&a[i].old);
                  a[i].num=i;
            }
            sort(a+1,a+1+N,cmp);
            int last=-1;
            for(int i=1;i<=N;i++)
            {
                  if(a[i].old!=last)
                  {
                        last=a[i].old;
                        a[i].v=++kind;
                  }
                  else a[i].v=kind;
            }
            sort(a+1,a+1+N,cmp1);
            for(int i=N;i>=1;i--)
            {
                  ans+=getsum(a[i].v);
                  add(a[i].v+1,1);
            }
            printf("%I64d\n",ans);
      }
	  //system("pause");
      return 0;
}

 

转载注明出处： http://blog.youkuaiyun.com/u011400953