题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4911
Inversion
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 2744 Accepted Submission(s): 1015
Problem Description
bobo has a sequence a1,a2,…,an. He is allowed to swap two adjacent numbers for no more than k times.
Find the minimum number of inversions after his swaps.
Note: The number of inversions is the number of pair (i,j) where 1≤i<j≤n and ai>aj.
Input
The input consists of several tests. For each tests:
The first line contains 2 integers n,k (1≤n≤105,0≤k≤109). The second line contains n integers a1,a2,…,an (0≤ai≤109).
Output
For each tests:
A single integer denotes the minimum number of inversions.
Sample Input
3 1
2 2 1
3 0
2 2 1
Sample Output
1
2
求交换k次以后所获得的数列的最小的逆序数。
如果逆序数大于0，说明数列中有两个数可以交换，使得逆序数-1。所以所求交换k次所得数列最小逆序数的结果就是排序结束后交换次数-k，这个结果大于等于0。
代码如下：
1 #include <iostream>
2 #include <cstring>
3 #include <cstdio>
4
5 using namespace std;
6
7 typedef long long LL;
8 const int maxn = 100010;
9 int num[maxn];
10 int Right[maxn], Left[maxn];
11 LL ans;
12
13 void merge(int* num, int p, int q, int r) {
14 int n1, n2, i, j, k;
15 n1 = q - p + 1;
16 n2 = r - q;
17 for(i = 0; i < n1; i++) {
18 Left[i] = num[p+i];
19 }
20 for(i = 0; i < n2; i++) {
21 Right[i] = num[q+i+1];
22 }
23 Left[n1] = Right[n2] = 0x7FFFFFFF;
24 i = 0;
25 j = 0;
26 for(k = p; k <= r; k++) {
27 if(Left[i] <= Right[j]) {
28 num[k] = Left[i];
29 i++;
30 }
31 else {
32 num[k] = Right[j];
33 j++;
34 ans += (n1 - i);
35 }
36 }
37 }
38
39 void mergesort(int* num, int p, int r) {
40 int q;
41 if(p < r) {
42 q = (p + r) / 2;
43 mergesort(num, p, q);
44 mergesort(num, q+1, r);
45 merge(num, p, q, r);
46 }
47 }
48
49 int main() {
50 int n, k;
51 while(~scanf("%d %d", &n, &k)) {
52 ans = 0;
53 for(int i = 0; i < n; i++) {
54 scanf("%d", &num[i]);
55 }
56 mergesort(num, 0, n-1);
57 cout << ((ans-k) > LL(0) ? ans-k : 0) << endl;
58 }
59 }