HDU 1394 树状数组求逆序对

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13036    Accepted Submission(s): 7968

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

 

题目意思：

给一个n，然后一个长度为n的数组，每个元素小于n且不重复，每次操作把首位元素放到末尾，求n次操作中最小的逆序对数目。

 

思路：

先求一下原始数组的逆序对数目，每次把首位放末尾的时候，num（逆序对数目）加上大于a[i]的，然后减去小于a[i]的。

 

代码：

 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <iostream>
 #include <vector>
 #include <queue>
 #include <cmath>
 #include <set>
 using namespace std;

 #define N 5005
 #define ll root<<1
 #define rr root<<1|1
 #define mid (a[root].l+a[root].r)/2

 int max(int x,int y){return x>y?x:y;}
 int min(int x,int y){return x<y?x:y;}
 int abs(int x,int y){return x<?-x:x;}

 int n;
 int b[N];
 int a[N];

 int lowbit(int x){
     return x&(-x);
 }

 void solve(int val,int x){
     while(x<=n){
         a[x]+=val;
         x+=lowbit(x);
     }
 }

 int sum(int x){
     int ans=;
     while(x>){
         ans+=a[x];
         x-=lowbit(x);
     }
     return ans;
 }

 main()
 {
     int t, i, j, k;
     while(scanf("%d",&n)==){
         memset(a,,sizeof(a));
         int ans=;
         for(i=;i<=n;i++) {
             scanf("%d",&b[i]);
             b[i]++;
             ans+=sum(n)-sum(b[i]);
             solve(,b[i]);
         }
         int num=ans;
         for(i=;i<=n;i++){
             num+=n-b[i]-b[i]+;//sum(n)-sum(b[i])-sum(b[i]-1);
             ans=min(ans,num);
         }
         printf("%d\n",ans);
     }
 }