poj 2299 逆序数树状数组-优快云博客

本文介绍了一种名为Ultra-QuickSort的排序算法，并详细解释了如何通过交换相邻元素将序列排序为升序的过程。此外，还提供了一个C++实现示例，用于计算将输入序列排序所需的最小交换次数。

（1)用结构体存每个数的数置和它的值，struct cnode{int value,int pos};

(2)pos位置取负再加上ＭＡＸＮ＋１，ＭＡＸＮ（是元素位置的最大值），

（３）排序：先按value从小到大排，再按pos从小到大排，剩下的就是求每个元素左下角的元素个数，交给树状数组来处理是再好不过的事了。

#include <iostream> #include <cstdio> #include <algorithm> #include <cstring> using namespace std; const int MAXN=500000; struct cnode { int pos; int value; }node[MAXN+10]; int tree[MAXN+10]; bool cmp(const cnode &e1,const cnode &e2) { if(e1.value!=e2.value) { return e1.value<e2.value? true:false; } else { return e1.pos<e2.pos? true:false; } } int n; int sum(int x) { int value=0; while(x>0) { value+=tree[x]; x-=(x&(-x)); } return value; } void insert(int x) { while(x<=MAXN) { tree[x]++; x+=(x&(-x)); } } int main() { //int t; //freopen("tree.out","w",stdout); // freopen("1804.in","r",stdin); // scanf("%d",&t); int num[MAXN+10]; int sub; int i; int Case=0; // while(t--) // { while(scanf("%d",&n)!=EOF) { if(n==0) { break; } memset(tree,0,sizeof(tree)); Case++; // scanf("%d",&n); for(i=1;i<=n;i++) { scanf("%d",&num[i]); } sub=1; for(i=n;i>0;i--) { node[sub].pos=-i+MAXN+1; node[sub].value=num[i]; sub++; } sort(&node[1],&node[n+1],cmp); long long ans=0; int samenum=0; node[0].value=-1000000-1; for(i=1;i<=n;i++) { if(node[i].value!=node[i-1].value) { samenum=0; ans+=sum(node[i].pos-1); //insert(node[i].pos); //ans+=beforevalue; } else { samenum++; ans+=sum(node[i].pos-1)-samenum; // insert(node[i].pos); } insert(node[i].pos); } // printf("Scenario #%d:/n%d/n/n",Case,ans); printf("%lld/n",ans); } return 0; }

题目：

Ultra-QuickSort

Time Limit: 7000MS		Memory Limit: 65536K
Total Submissions: 20199		Accepted: 7146

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