标签:include sequence LL QuickSort 离散 input Ultra
题目链接:http://poj.org/problem?id=2299
Ultra-QuickSortTime Limit: 7000MS | Memory Limit: 65536K | |
Total Submissions: 75831 | Accepted: 28402 |
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 sequenceUltra-QuickSort produces the output
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 题意:找到给定数组中有多少逆序对 思路:树状数组离散化,存一个找一次 不需要全部存进去再找(这样是找不出来的) 存的时候就找,这样就知道有多少个是逆序的了 代码:#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> #include<vector> using namespace std; typedef long long LL; const LL mod=1e9+7; const LL INF=1e9+7; const int maxn=5e5+50; LL a[maxn],b[maxn],c[maxn]; LL len; LL lowbit(LL x) { return x&(-x); } LL query(LL x) { LL ret=0; while(x>0) { ret+=c[x]; x-=lowbit(x); } return ret; } void update(LL x) { while(x<=500000) { c[x]++; x+=lowbit(x); } } int main() { LL N; while(scanf("%lld",&N)!=EOF) { if(N==0) break; memset(a,0,sizeof(a)); memset(b,0,sizeof(b)); memset(c,0,sizeof(c)); LL ans=0; for(int i=1;i<=N;i++) { scanf("%lld",&a[i]); b[i]=a[i]; } sort(b+1,b+N+1); len=unique(b+1,b+N+1)-b; for(int i=1;i<=N;i++) { // cout<<"*"<<endl; LL x=lower_bound(b+1,b+len+1,a[i])-b;//找到a[i]在b数组中所在位置 ans+=i-query(x-1)-1;//query找在x之前的数有多少个 这些都是顺序对的 update(x); } printf("%lld\n",ans); } return 0; }
标签:include,sequence,LL,QuickSort,离散,input,Ultra
来源: https://www.cnblogs.com/caijiaming/p/10896713.html
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