POJ-2299 Ultra-QuickSort(逆序对个数)

Posted 2022-09-16 MichaelZona

Ultra-QuickSort

Time Limit: 7000MS		Memory Limit: 65536K
Total Submissions: 57461		Accepted: 21231

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​​

 

哼你以为你让我WA我就不发日志了吗QAQ

 

经过快一年的努力终于改对了！！哈哈哈哈哈

 

问题出在离散化，相等的数我把弄了不同的位置所以错了

1 #include "bits/stdc++.h"
 2 #define mem(a,b) memset(a,b,sizeof(a))
 3 using namespace std;
 4 typedef long long LL;
 5 const int MAX=500005;
 6 int n;
 7 struct Poi
 8     int w,p,z;
 9 cc[MAX];
10 int c[MAX];
11 void add(int x,int y)for (;x<=n;x+=(x&-x)) c[x]+=y;
12 int search(int x)int an(0);for (;x>0;x-=(x&-x)) an+=c[x];return an;
13 bool cmp1(Poi x,Poi y)return x.w>y.w;
14 bool cmp2(Poi x,Poi y)return x.p<y.p;
15 void init()
16     int i,j;
17     for (i=1;i<=n;i++)
18         scanf("%d",&cc[i].w);
19         cc[i].p=i;
20     
21     sort(cc+1,cc+n+1,cmp1);
22     for (i=1;i<=n;i++)

  if (vv[i].w!=vv[i-1].w)

        vv[i].z=++tt;

           else

   vv[i].z=tt;         24     
25     sort(cc+1,cc+n+1,cmp2);
26     mem(c,0);
27 
28 int main()
29     freopen ("ultra.in","r",stdin);
30     freopen ("ultra.out","w",stdout);
31     int i,j;
32     int ans;
33     while (scanf("%d",&n),n)
34         init();ans=0;
35         for (i=1;i<=n;i++)
36             ans+=search(cc[i].z-1);
37             add(cc[i].z,1);
38         
39         printf("%d\\n",ans);
40     
41     return 0;
42