hdu 4911 Inversion（歸併排序求逆序對數）

標籤：歸併排序

Inversion                                                                            Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)

Problem Descriptionbobo 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. 
InputThe 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). 
OutputFor each tests:

A single integer denotes the minimum number of inversions. 
Sample Input

3 12 2 13 02 2 1

 
Sample Output

12

 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];//歸併排序的合併作業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);    }}