HDU 4911 Inversion (merge sort for reverse order)

InversionTime limit:2000/1000 MS (java/others) Memory limit:131072/131072 K (java/others)
Total submission (s): 1898 Accepted Submission (s): 743


Problem Descriptionbobo has a sequence a1,a2,...,. He is allowed to swap adjacentNumbers 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

#include <iostream> #include <string.h>using namespace std; #define MAX 100010int aux[max];int a[max];int N;    Long long K;long long ans=0;void merge (int a[],int l,int mid,int h) {int i=l;    int j=mid+1;    for (int k=l;k<=h;++k) {aux[k]=a[k];        } for (int k=l;k<=h;k++) {if (i>mid) a[k]=aux[j++];        else if (j>h) a[k]=aux[i++];        else if (Aux[i]<=aux[j]) a[k]=aux[i++];            else {a[k]=aux[j++];        ans+= (mid-i+1);    }}}void mergesort (int a[],int l,int h) {if (h<=l) return;    int mid=l+ (H-L)/2;    MergeSort (A,l,mid);    MergeSort (A,MID+1,H); Merge (a,l,mid,h);}    int main (int argc, char *argv[]) {//freopen ("4911.in", "R", stdin);        while (scanf ("%d%lld", &n,&k)!=eof) {ans=0;        memset (aux,0,sizeof (aux));        for (int k=0;k<n;++k) scanf ("%d", &a[k]);        MergeSort (a,0,n-1);        if (K<ans) printf ("%lld\n", ans-k);    else printf ("0\n"); } rEturn 0;} 

Remember to define ANS and k as Long long

HDU 4911 Inversion (merge sort for reverse order)