POJ 2299 Ultra-QuickSort [树状数组做法]

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

Source

Waterloo local 2005.02.05

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求逆序对的题，可以用归并做，也可以用树状数组~

树状数组+离散化~

因为n<=50000，而每个数<=999999999，如果按数本身来存会爆掉，所以离散化一下用标号来存~

神奇的离散化方法，本蒟蒻还是第一次用呢……

#include<cstdio>#include<algorithm>using namespace std;int n,c[500005],reflect[500005];long long tot;struct node{int val,num;}a[500005];bool cmp(node u,node v){return u.val<v.val;}int lowbit(int u){return u&(-u);}void add(int u){for(int i=u;i<=n;i+=lowbit(i)) c[i]++;}int getnum(int u){int ans=0;for(int i=u;i;i-=lowbit(i)) ans+=c[i];return ans;}int main(){while(scanf("%d",&n)!=EOF && n){tot=0;for(int i=1;i<=n;i++){scanf("%d",&a[i].val);a[i].num=i;}sort(a+1,a+n+1,cmp);for(int i=1;i<=n;i++) reflect[a[i].num]=i;for(int i=1;i<=n;i++) c[i]=0;for(int j=1;j<=n;j++){add(reflect[j]);tot+=(j-getnum(reflect[j]));}printf("%lld\n",tot);}return 0;}