hdu 1394 Minimum Inversion Number

最新推荐文章于 2024-05-07 07:50:01 发布

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原文链接：http://www.cnblogs.com/GadyPu/p/4538836.html

本文介绍了一道算法题——最小逆序数的求解方法。通过归并排序算法实现，详细展示了如何计算一系列整数排列中逆序对的最小数量，并提供了完整的C++代码示例。

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题目连接

http://acm.hdu.edu.cn/showproblem.php?pid=1394

Minimum Inversion Number

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

归并排序求逆序数。。

 1 #include<algorithm>
 2 #include<iostream>
 3 #include<cstdlib>
 4 #include<cstring>
 5 #include<cstdio>
 6 #define mid ((l+r)>>1)
 7 using std::min;
 8 const int Max_N = 5010;
 9 const int INF = ~0u >> 1;
10 int cnt, arr[Max_N], ret[Max_N], temp[Max_N];
11 void Merge(int *A, int l, int m, int r) {
12     int p = 0;
13     int x = l, y = m + 1;
14     while (x <= m && y <= r) {
15         if (A[x] > A[y]) cnt += m - x + 1, temp[p++] = A[y++];
16         else temp[p++] = A[x++];
17     }
18     while (x <= m) temp[p++] = A[x++];
19     while (y <= r) temp[p++] = A[y++];
20     for (x = 0; x < p; x++) A[l + x] = temp[x];
21 }
22 void MergeSort(int *A, int l, int r) {
23     if (l < r) {
24         MergeSort(A, l, mid);
25         MergeSort(A, mid + 1, r);
26         Merge(A, l, mid, r);
27     }
28 }
29 int main() {
30 #ifdef LOCAL
31     freopen("in.txt", "r", stdin);
32     freopen("out.txt", "w+", stdout);
33 #endif
34     int n;
35     while (~scanf("%d", &n)) {
36         for (int i = 0; i < n; i++) scanf("%d", &arr[i]), ret[i] = arr[i];
37         cnt = 0;
38         MergeSort(arr, 0, n - 1);
39         int res = INF;
40         for (int i = 0; i < n; i++) {
41             cnt = cnt + n - ret[i] - ret[i] - 1;
42             res = min(res, cnt);
43         }
44         printf("%d\n", res);
45     }
46     return 0;
47 }