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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
Sample Output
Author
CHEN, Gaoli
Source
Recommend
Ignatius.L
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 }
[hdu1394]Minimum Inversion Number(树状数组)
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原文地址：http://www.cnblogs.com/Pumbit-Legion/p/5874029.html
