hdu 1394 Minimum Inversion Number - 树状数组

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

---

　　题目大意 求循环排列最小逆序对数。

　　老是刷cf，不做暑假作业估计会被教练鄙视，所以还是做做暑假作业。

　　先用各种方法算出原始序列的逆序对数。

　　显然你可直接算出把开头的一个数挪到序列后面增加的逆序对数。(后面有多少个比它大减去有多少个比它小)

　　于是这道题就水完了。

Code

 /**
  * hdu
  * Problem#1394
  * Accepted
  * Time: 62ms
  * Memory: 1680k
  */
 #include <iostream>
 #include <fstream>
 #include <sstream>
 #include <cstdio>
 #include <ctime>
 #include <cmath>
 #include <cctype>
 #include <cstring>
 #include <cstdlib>
 #include <algorithm>
 #include <map>
 #include <set>
 #include <list>
 #include <stack>
 #include <queue>
 #include <vector>
 #ifndef WIN32
 #define Auto "%lld"
 #else
 #define Auto "%I64d"
 #endif
 using namespace std;
 typedef bool boolean;
 const signed int inf = (signed)((1u << ) - );
 const signed long long llf = (signed long long)((1ull << ) - );
 const double eps = 1e-;
 const int binary_limit = ;
 #define smin(a, b) a = min(a, b)
 #define smax(a, b) a = max(a, b)
 #define max3(a, b, c) max(a, max(b, c))
 #define min3(a, b, c) min(a, min(b, c))
 template<typename T>
 inline boolean readInteger(T& u){
     char x;
     int aFlag = ;
     while(!isdigit((x = getchar())) && x != '-' && x != -);
     if(x == -) {
         ungetc(x, stdin);
         return false;
     }
     if(x == '-'){
         x = getchar();
         aFlag = -;
     }
     for(u = x - ''; isdigit((x = getchar())); u = (u << ) + (u << ) + x - '');
     ungetc(x, stdin);
     u *= aFlag;
     return true;
 }

 #define lowbit(x) ((x) & (-x))

 typedef class IndexedTree {
     public:
         int* lis;
         int s;
         IndexedTree() {        }
         IndexedTree(int s):s(s) {
             lis    = new int[(s + )];
             memset(lis, , sizeof(int) * (s + ));
         }

         inline void add(int idx, int val) {
             for(; idx <= s; idx += lowbit(idx))
                 lis[idx] += val;
         }

         inline int getSum(int idx) {
             int ret = ;
             for(; idx; idx -= lowbit(idx))
                 ret += lis[idx];
             return ret;
         }
 }IndexedTree;

 int n;
 int* arr;
 inline boolean init() {
     if(!readInteger(n))    return false;
     arr = new int[(n + )];
     for(int i = ; i <= n; i++)
         readInteger(arr[i]), arr[i] += ;
     return true;
 }

 IndexedTree it;
 int ans, cmp;
 inline void solve() {
     ans = , cmp = ;
     it = IndexedTree(n);
     for(int i = ; i <= n; i++) {
         it.add(arr[i], );
 //        cout << 1;
         cmp += i - it.getSum(arr[i]);
     }
     ans = cmp;
     for(int i = ; i < n; i++) {
         cmp -= arr[i] - , cmp += n - arr[i];
         smin(ans, cmp);
     }
     printf("%d\n", ans);
 }

 inline void clear() {
     delete[] arr;
     delete[] it.lis;
 }

 int main() {
     while(init()) {
         solve();
         clear();
     }
     return ;
 }