HDU 2689 Sort it【树状数组】
Sort it
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 4672 Accepted Submission(s): 3244
Problem Description
You want to processe a sequence of n distinct integers by swapping two adjacent sequence elements until the sequence is sorted in ascending order. Then how many times it need. For example, 1 2 3 5 4, we only need one operation : swap 5 and 4.
Input
The input consists of a number of test cases. Each case consists of two lines: the first line contains a positive integer n (n <= 1000); the next line contains a permutation of the n integers from 1 to n.
Output
For each case, output the minimum times need to sort it in ascending order on a single line.
Sample Input
3
1 2 3
4
4 3 2 1
Sample Output
0
6
Author
WhereIsHeroFrom
Source
ZJFC 2009-3 Programming Contest
Recommend
题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2689
分析：好吧，我智障了，听说有种很快的写法，但好像跑的速度没我快？应该是我代码跑的最快吧，写法也是最复杂的，这题我是用树状数组乱搞的，反正15ms，相当快啊！
下面给出AC代码：
1 #include <bits/stdc++.h>
2 using namespace std;
3 typedef long long ll;
4 inline int read()
5 {
6 int x=0,f=1;
7 char ch=getchar();
8 while(ch<'0'||ch>'9')
9 {
10 if(ch=='-')
11 f=-1;
12 ch=getchar();
13 }
14 while(ch>='0'&&ch<='9')
15 {
16 x=x*10+ch-'0';
17 ch=getchar();
18 }
19 return x*f;
20 }
21 inline void write(int x)
22 {
23 if(x<0)
24 {
25 putchar('-');
26 x=-x;
27 }
28 if(x>9)
29 {
30 write(x/10);
31 }
32 putchar(x%10+'0');
33 }
34 const int N=1001;
35 int n;
36 int num[N];
37 int c[N];
38 struct node
39 {
40 int x;
41 int id;
42 }q[N] ;
43 bool cmp(node a,node b)
44 {
45 return a.x<b.x;
46 }
47 int lowbit(int x)
48 {
49 return x&(-x);
50 }
51 int getsum(int x)
52 {
53 int s=0;
54 while(x>0)
55 {
56 s+=c[x];
57 x-=lowbit(x);
58 }
59 return s;
60 }
61 void add(int x,int y)
62 {
63 while(x<=n)
64 {
65 c[x]+=y;
66 x+=lowbit(x);
67 }
68 }
69 int main()
70 {
71 while(scanf("%d",&n)!=EOF)
72 {
73 memset(c,0,sizeof(c));
74 memset(num,0,sizeof(num));
75 for(int i=1;i<=n;i++)
76 {
77 scanf("%d",&q[i].x);
78 q[i].id=i;
79 }
80 sort(q+1,q+1+n,cmp);
81 for(int i=1;i<=n;i++)
82 {
83 num[q[i].id]=i;
84 }
85 int sum = 0;
86 for(int i=1;i<=n;i++)
87 {
88 add(num[i],1);
89 sum+=getsum(n)-getsum(num[i]);
90 }
91 printf("%d\n",sum);
92 }
93 return 0;
94 }腾讯云开发者
