poj-----Ultra-QuickSort(离散化+树状数组)

Ultra-QuickSort

Time Limit: 7000MS		Memory Limit: 65536K
Total Submissions: 38258		Accepted: 13784

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

Sample Output

6
0

Source

Waterloo local 2005.02.05

代码：

 1 //离散化+树状数组
 2 #include<stdio.h>
 3 #include<stdlib.h>
 4 #include<string.h>
 5 #define maxn 500000
 6 struct node
 7 {
 8     int val;
 9     int id;
10 };
11 node stu[maxn+5];
12 int n;
13 __int64 cnt;
14 int bb[maxn+5];
15 int lowbit(int x)
16 {
17   return x&(-x);
18 }
19 void ope(int x)
20 {
21     while(x<=n)
22     {
23      bb[x]++;
24      x+=lowbit(x);
25     }
26 }
27 int sum(int x)
28 {
29     int ans=0;
30     while(x>0)
31     {
32      ans+=bb[x];
33      x-=lowbit(x);
34     }
35     return ans;
36 }
37 int cmp(const void *a ,const void *b)
38 {
39   return (*(node *)a).val -(*(node *)b).val;
40 }
41 int main()
42 {
43     int i;
44     while(scanf("%d",&n),n)
45     {
46         memset(bb,0,sizeof(int)*(n+1));
47         for(i=0;i<n;i++)
48         {
49          scanf("%d",&stu[i].val);
50          stu[i].id=i+1;
51         }
52         qsort(stu,n,sizeof(stu[0]),cmp);
53         cnt=0;
54         for(i=0;i<n;i++)
55         {
56             cnt+=sum(n)-sum(stu[i].id);
57             ope(stu[i].id);
58         }
59         printf("%I64dn",cnt);
60     }
61   return 0;
62 }

归并排序：

代码：

 1 //归并排序求逆序数
 2 #include<stdio.h>
 3 #include<string.h>
 4 #include<stdlib.h>
 5 #define maxn 500000
 6 int cc[maxn+5];
 7 int aa[maxn+5];
 8 __int64 cnt;
 9 void merge(int low ,int mid,int hig)
10 {
11     int i,j,k;
12     i=low;
13     j=mid;
14     k=0;
15     while(i<mid&&j<hig)
16     {
17         if(aa[i]>aa[j])
18         {
19             cc[k++]=aa[j++];
20             cnt+=mid-i;
21         }
22         else
23             cc[k++]=aa[i++];
24     }
25     while(i<mid)
26         cc[k++]=aa[i++];
27     while(j<hig)
28         cc[k++]=aa[j++];
29     for(k=0,i=low;i<hig;i++)
30         aa[i]=cc[k++];
31 }
32 void merge_sort(int st,int en)
33 {
34     int mid;
35     if(st+1<en)
36     {
37         mid=st+(en-st)/2;
38         merge_sort(st,mid);
39         merge_sort(mid,en);
40         merge(st,mid,en);
41     }
42 }
43 int main()
44 {
45     int n,i;
46     while(scanf("%d",&n),n)
47     {
48         cnt=0;
49         for(i=0;i<n;i++)
50          scanf("%d",aa+i);
51          merge_sort(0,n);
52          printf("%I64dn",cnt);
53     }
54     return 0;
55 }

非递归归并排序：代码：

 1 //归并排序求逆序数
 2 #include<stdio.h>
 3 #include<string.h>
 4 #include<stdlib.h>
 5 #define maxn 500000
 6 int cc[maxn+5];
 7 int aa[maxn+5];
 8 __int64 cnt;
 9 void merge(int low ,int mid,int hig)
10 {
11     int i,j,k;
12     i=low;
13     j=mid;
14     k=0;
15     while(i<mid&&j<hig)
16     {
17         if(aa[i]>aa[j])
18         {
19             cc[k++]=aa[j++];
20             cnt+=mid-i;
21         }
22         else
23             cc[k++]=aa[i++];
24     }
25     while(i<mid)
26         cc[k++]=aa[i++];
27     while(j<hig)
28         cc[k++]=aa[j++];
29     for(k=0,i=low;i<hig;i++)
30         aa[i]=cc[k++];
31 }
32 //void merge_sort(int st,int en)
33 //{
34 //    int mid;
35 //    if(st+1<en)
36 //    {
37 //        mid=st+(en-st)/2;
38 //        merge_sort(st,mid);
39 //        merge_sort(mid,en);
40 //        merge(st,mid,en);
41 //    }
42 //}
43 void merge_sort(int st,int en)
44 {
45     int s,t,i;
46     t=1;
47     while(t<=(en-st))
48     {
49         s=t;
50         t<<=1;
51         i=st;
52         while(i+t<=en)
53         {
54             merge(i,i+s,i+t);
55             i+=t;
56         }
57         if(i+s<en)
58             merge(i,i+s,en);
59     }
60       if(i+s<en)
61             merge(i,i+s,en);
62 }
63 int main()
64 {
65     int n,i;
66     while(scanf("%d",&n),n)
67     {
68         cnt=0;
69         for(i=0;i<n;i++)
70          scanf("%d",aa+i);
71          merge_sort(0,n);
72          printf("%I64dn",cnt);
73     }
74     return 0;
75 }