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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;
}