HDU Problem 1394 Minimum Inversion Number【树状数组】-白红宇

HDU Problem 1394 Minimum Inversion Number【树状数组】

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发布时间：2019-06-27

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Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 18277 Accepted Submission(s): 11093

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

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一开始没有看到数组是连续的。。。。

先求出最原始的逆序数和ans

如果数组是连续的那就好办了，对于第一个元素，有a【0】 - 1 个元素是小于它的，而有n - a【0】‘个元素是大于它的。所以每次转换就变成了

ans = ans - a【0】+ 1 + n + - a【0】；

#include 
     
      #include 
      
       #include 
       #include 
        
         #include 
         
          #include 
          
           #include 
           
            #define space " "using namespace std;const int MAXN = 5000 + 10;const int INF = 0x3f3f3f3f;int n, ar[MAXN], b[MAXN], num[MAXN];int sum(int x) { int s = 0; while (x >= 1) { s += num[x]; x -= x&-x; } return s;}void add(int x, int y) { while (x <= n) { num[x] += y; x += x&-x; }}int main() { while (scanf("%d", &n) != EOF) { for (int i = 1; i <= n; i++) { scanf("%d", &ar[i]); ar[i]++; } int ans = 0; memset(num, 0, sizeof(num)); for (int i = 1; i <= n; i++) { add(ar[i], 1); ans += i - sum(ar[i]); } int minn = ans; for (int i = 1; i <= n; i++) { ans += n - ar[i] - ar[i] + 1; minn = min(minn, ans); } printf("%d\n", minn); } return 0;}

转载于:https://www.cnblogs.com/cniwoq/p/6770814.html

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