Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 18395    Accepted Submission(s): 11168

Problem Description

The inversion number of a given number sequence a1, a2, ..., an is the number of pairs (ai, aj) that satisfy i < j and ai > aj.

For a given sequence of numbers a1, a2, ..., an, if we move the first m >= 0 numbers to the end of the seqence, we will obtain another sequence. There are totally n such sequences as the following:

a1, a2, ..., an-1, an (where m = 0 - the initial seqence)
a2, a3, ..., an, a1 (where m = 1)
a3, a4, ..., an, a1, a2 (where m = 2)
...
an, a1, a2, ..., an-1 (where m = n-1)

You are asked to write a program to find the minimum inversion number out of the above sequences.

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 <= 5000); the next line contains a permutation of the n integers from 0 to n-1.

Output

For each case, output the minimum inversion number on a single line.

Sample Input

10
1 3 6 9 0 8 5 7 4 2

Sample Output

16

Author

CHEN, Gaoli

Source

ZOJ Monthly, January 2003

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更更更

隔壁YNY看到就直接暴力，当然T了(哈哈哈一起笑他)

动动脑子，题目说是一个环

那么每转一次，就相当于把第一个数放到最后面

考虑第一个数对原有答案的贡献是a[1]-1，也就是小于它的个数(数据是1到n的排列)

最后一个数对原答案的贡献相反

那么移动后当前逆序对数就要减去比第一个数小的个数，再加上比它大的数的个数

这样我们求出一次移动后的逆序对数

这时候我们发现下一次移动直接修改答案就好，不需要改动树状数组

推出式子ans+=n-a[i]-(a[i]-1)

 1 #include<stdio.h>
 2 #include<stdlib.h>
 3 #include<string.h>
 4 #define LL long long
 5 int bit[5050]={0},n,a[5050];
 6 inline LL min(LL a,LL b){
 7     return a<b?a:b;
 8 }
 9 inline int lb(int x){
10     return x&(-x);
11 }
12 inline LL q(int x){
13     LL ans=0;
14     while(x){
15         ans+=bit[x];
16         x-=lb(x);
17     }
18     return ans;
19 }
20 inline int c(int x){
21     while(x<=n){
22         bit[x]++;
23         x+=lb(x);
24     }
25     return 0;
26 }
27 int main(){
28     while(scanf("%d",&n)!=EOF){
29         memset(bit,0,sizeof(bit));
30         LL ans=0;
31         for(int i=1;i<=n;i++){
32             scanf("%d",&a[i]);
33             a[i]++;
34             ans+=q(n)-q(a[i]);
35             c(a[i]);
36         }
37         LL mn=ans;
38         mn=min(mn,ans);
39         for(int i=1;i<=n;i++){
40             ans+=n-a[i]-(a[i]-1);
41             mn=min(mn,ans);
42         }
43         printf("%lld\n",mn);
44     }
45
46     return 0;
47 }