Inversion
Time Limit: 2000/1000 MS (Java/Others) Memory Limit:
131072/131072 K (Java/Others)
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
题意:给出n个数,每次能够交换相邻的两个数。最多交换k次,求交换后最小的逆序数是多少。
分析:假设逆序数大于0。则存在1 ≤ i < n,使得交换ai和ai+1后逆序数减1。所以最后的答案就是max((inversion-k), 0)。
利用归并排序求出原序列的逆序对数就能够解决这个问题了。
#include<stdio.h> #include<string.h> #define N 100005 __int64 cnt, k; int a[N],c[N]; //归并排序的合并操作 void merge(int a[], int first, int mid, int last, int c[]) { int i = first, j = mid + 1; int m = mid, n = last; int k = 0; while(i <= m || j <= n) { if(j > n || (i <= m && a[i] <= a[j])) c[k++] = a[i++]; else { c[k++] = a[j++]; cnt += (m - i + 1); } } for(i = 0; i < k; i++) a[first + i] = c[i]; } //归并排序的递归分解和合并 void merge_sort(int a[], int first, int last, int c[]) { if(first < last) { int mid = (first + last) / 2; merge_sort(a, first, mid, c); merge_sort(a, mid+1, last, c); merge(a, first, mid, last, c); } } int main() { int n; while(~scanf("%d%I64d",&n,&k)) { memset(c, 0, sizeof(c)); cnt = 0; for(int i = 0; i < n; i++) scanf("%d", &a[i]); merge_sort(a, 0, n-1, c); if(k >= cnt) cnt = 0; else cnt -= k; printf("%I64d\n",cnt); } }