HDU4911：Inversion

標籤：hdu   歸併   

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根據逆序數的定理如果逆序數大於0，那麼必定存在1<=i<n使得i和i+1交換後逆序數減1假設原逆序數為cnt，這樣的話，我們就可以得到答案是max（cnt-k,0）求逆序數可以用歸併的方法
#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;int a[100005];int left[100005], right[100005];__int64 cnt;void merge(int* a, int p, int q, int r){    int i, j, k, n1, n2;    n1 = q-p+1;    n2 = r-q;    for (i=0; i<n1; i++)    {        left[i] = a[p+i];    }    for (i=0; i<n2; i++)    {        right[i] = a[q+i+1];    }    left[n1] = right[n2] = 0x7fffffff;    i = j = 0;    for (k=p; k<=r; k++)    {        if (left[i] <= right[j])        {            a[k] = left[i];            i++;        }        else        {            a[k] = right[j];            j++;            cnt += n1-i; /**此步驟是在歸併排序法中加的一句，用來計數求逆序數的數目**/        }    }    return;}void mergesort(int* a, int p, int r){    int q;    if (p < r)    {        q = (p+r)/2;        mergesort(a, p, q);        mergesort(a, q+1, r);        merge(a, p, q, r);    }    return ;}int main(){    int n,k,i,j;    while(~scanf("%d%d",&n,&k))    {        cnt = 0;        for(i = 0;i<n;i++)        scanf("%d",&a[i]);        mergesort(a,0,n-1);        printf("%I64d\n",max(cnt-k,(__int64)0));    }    return 0;}