Poj 2299 - Ultra-QuickSort 离散化,树状数组,逆序对

Ultra-QuickSort Time Limit: 7000MS   Memory Limit: 65536K Total Submissions: 52306   Accepted: 19194 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   题意：给定n个数，只能交换相邻的两个元素，至少交换几次，成为递增序列。 题解： 明显要求序列的逆序对数目。。。 对于样例： 5 9 1 0 5 4 我们将其排序： 0 1 4 5 9 在每个位置上初始放为1. 1 1 1 1 1 然后，从原序列开始遍历。 先到9，我们把其排好序的位置拿出， 即为5。 然后统计位置5之前有多少1。 1 1 1 1 1 ————  > 4个  ans+=4 然后把5号位置放为0。 1 1 1 1 0 继续操作即可。。。 这个树状数组维护即可。。。 注意开long long和原序列排序后要去重。。。 1 #include<bits/stdc++.h> 2 using namespace std; 3 #define MAXN 500010 4 #define LL long long 5 int n,BIT[MAXN],a[MAXN],aa[MAXN]; 6 int read() 7 { 8 int s=0,fh=1;char ch=getchar(); 9 while(ch<'0'||ch>'9'){if(ch=='-')fh=-1;ch=getchar();} 10 while(ch>='0'&&ch<='9'){s=s*10+(ch-'0');ch=getchar();} 11 return s*fh; 12 } 13 int Lowbit(int o){return o&(-o);} 14 void Update(int o,int o1) 15 { 16 while(o<=n) 17 { 18 BIT[o]+=o1; 19 o+=Lowbit(o); 20 } 21 } 22 int Sum(int o) 23 { 24 int sum=0; 25 while(o>0) 26 { 27 sum+=BIT[o]; 28 o-=Lowbit(o); 29 } 30 return sum; 31 } 32 int main() 33 { 34 int tot,i,wz; 35 LL ans; 36 while(1) 37 { 38 n=read();if(n==0)break; 39 for(i=1;i<=n;i++){a[i]=read();aa[i]=a[i];} 40 memset(BIT,0,sizeof(BIT)); 41 sort(a+1,a+n+1); 42 tot=unique(a+1,a+n+1)-(a+1); 43 for(i=1;i<=tot;i++)Update(i,1); 44 ans=0; 45 for(i=1;i<=n;i++) 46 { 47 wz=lower_bound(a+1,a+tot+1,aa[i])-a; 48 ans+=(LL)Sum(wz-1); 49 if(BIT[wz]!=0)Update(wz,-1); 50 } 51 printf("%lld\n",ans); 52 } 53 fclose(stdin); 54 fclose(stdout); 55 return 0; 56 } View Code