Crossings

Time Limit: 2 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/gym/100463

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

HINT

题意

逆序对求解

题解：

树状数组水

代码

 1 #include <cstdio>
 2 #include <cmath>
 3 #include <cstring>
 4 #include <ctime>
 5 #include <iostream>
 6 #include <algorithm>
 7 #include <set>
 8 #include <vector>
 9 #include <queue>
10 #include <map>
11 #include <stack>
12 #define MOD 1000000007
13 #define maxn 32001
14 using namespace std;
15 typedef __int64 ll;
16 inline ll read()
17 {
18     ll x=0,f=1;
19     char ch=getchar();
20     while(ch<‘0‘||ch>‘9‘)
21     {
22         if(ch==‘-‘)f=-1;
23         ch=getchar();
24     }
25     while(ch>=‘0‘&&ch<=‘9‘)
26     {
27         x=x*10+ch-‘0‘;
28         ch=getchar();
29     }
30     return x*f;
31 }
32 //*******************************************************************
33
34 struct ss
35 {
36     int v,index;
37 } in[500005];
38 int c[500005];
39 int a[500005];
40 int n;
41 bool cmp(ss s1,ss s2)
42 {
43     return s1.v<s2.v;
44 }
45 int lowbit(int x)
46 {
47     return x&(-x);
48 }
49 int getsum(int x)
50 {
51     int sum=0;
52     while(x>0)
53     {
54         sum+=c[x];
55         x-=lowbit(x);
56     }
57     return sum;
58 }
59 void update(int x,int value)
60 {
61     while(x<=n)
62     {
63         c[x]+=value;
64         x+=lowbit(x);
65     }
66 }
67 int main()
68 {
69
70     while(scanf("%d",&n)!=EOF)
71     {
72         if(n==0) break;
73         memset(c,0,sizeof(c));
74         for(int i=1; i<=n; i++)
75         {
76             in[i].v=read();
77             in[i].index=i;
78         }
79         sort(in+1,in+n+1,cmp);
80         for(int i=1; i<=n; i++)a[in[i].index]=i; //离散化处理
81         ll ans=0;
82         for(int i=1; i<=n; i++)
83         {
84             update(a[i],1);
85             ans+=i-getsum(a[i]);
86         }
87         cout<<ans<<endl;
88     }
89
90     return 0;
91 }