hdu 4911 Inversion(歸併排序求逆序對數)2014多校訓練第5場
阿新 • • 發佈:2019-01-02
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
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 題意:給出n個數,每次可以交換相鄰的兩個數,最多交換k次,求交換後最小的逆序數是多少。 分析:如果逆序數大於0,則存在1 ≤ i < n,使得交換ai和ai+1後逆序數減1。所以最後的答案就是max((inversion-k), 0)。利用歸併排序求出原序列的逆序對數就可以解決問題了。
#include<stdio.h> #include<string.h> #define N 100005 __int64 cnt, k; int a[N],c[N]; //歸併排序的合併操作 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]; } //歸併排序的遞迴分解和合並 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); } }