poj2299 Ultra-QuickSort

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

59105431230

Sample Output

60

这道题可以用线段树做，也可以用树状数组做，这道题树状数组快了一倍啊。。题目求的就是整个排序的逆序数，可以先离散化，然后每插入一个数，就判断前面有几个数（即比它小的数的个数）sum[i]，然后在这个数前且比这个数大的数的个数为i-sum[i],把它们都加起来就行了。逆序数的定义：排在pi前面并且比pi大的元素的个数称为元素pi的逆序数。

线段树代码：

#include<iostream>#include<stdio.h>#include<string.h>#include<math.h>#include<vector>#include<map>#include<queue>#include<stack>#include<string>#include<algorithm>using namespace std;#define maxn 500010__int64 sum;struct node{int id,num,num1;}a[maxn];struct edge{int l,r,num;}b[4*maxn];bool cmp(node a,node b){return a.num<b.num;}bool cmp1(node a,node b){return a.id<b.id;}void build(int l,int r,int i){int mid;b[i].l=l;b[i].r=r;b[i].num=0;if(l==r)return;mid=(l+r)/2;build(l,mid,i*2);build(mid+1,r,i*2+1);}void question(int id,int i){int mid;if(b[i].l==b[i].r){b[i].num=1;return;}mid=(b[i].l+b[i].r)/2;if(id>mid)question(id,i*2+1);else {sum+=b[i*2+1].num;question(id,i*2);}b[i].num=b[i*2].num+b[i*2+1].num;}int main(){int n,m,i,j,c;while(scanf("%d",&n)!=EOF && n!=0){build(1,maxn,1);for(i=1;i<=n;i++){scanf("%d",&a[i].num);a[i].id=i;}sort(a+1,a+1+n,cmp);for(i=1;i<=n;i++){a[i].num1=i;}sort(a+1,a+1+n,cmp1);sum=0;for(i=1;i<=n;i++){question(a[i].num1,1);}printf("%I64d\n",sum);}return 0;}

树状数组代码：

#include<iostream>#include<stdio.h>#include<string.h>#include<math.h>#include<vector>#include<map>#include<queue>#include<stack>#include<string>#include<algorithm>using namespace std;#define maxn 500005struct node{int id,num;}a[maxn];int c[maxn];bool cmp1(node a,node b){return a.num<b.num;}bool cmp2(node a,node b){return a.id<b.id;}int lowbit(int x){return x&(-x);}void update(int pos,int num){while(pos<=maxn){c[pos]+=num;pos+=lowbit(pos);}}int sum(int pos){int num=0;while(pos>0){num+=c[pos];pos-=lowbit(pos);}return num;}int main(){int n,m,i,j;__int64 num;while(scanf("%d",&n)!=EOF && n!=0){for(i=1;i<=n;i++){scanf("%d",&a[i].num);a[i].id=i;}sort(a+1,a+1+n,cmp1);for(i=1;i<=n;i++) a[i].num=i;sort(a+1,a+1+n,cmp2);memset(c,0,sizeof(c));num=0;for(i=1;i<=n;i++){update(a[i].num,1);num+=i-sum(a[i].num);}printf("%I64d\n",num);}return 0;}