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题意:给一个长度为n的序列,问经过最多k次两两交换(仅限相邻的数之间),最少还有多少逆序对.

分析:归并求出总的逆序对数,然后ans=max(求出的总逆序对-k,0).

哇&#$%,Anything is possible.

参考代码:

#include<cstdio>#include<cmath>#include<cstring>#include<algorithm>#include<iostream>using namespace std;typedef long long LL;const int maxn = 1e5+10;int n;LL k;int a[maxn];int tmp[maxn];LL ans;//emmmmmm数据有点大 用LL才能过void Merge( int l, int mid, int r){    int i = l;    int j = mid+1;    int k = l;    while( i <= mid && j <= r)    {        if( a[i] > a[j])        {            tmp[k++] = a[j++];            ans += mid-i+1;        }        else            tmp[k++] = a[i++];    }    while( i <= mid)        tmp[k++] = a[i++];    while( j <= r)        tmp[k++] = a[j++];    for( int i = l; i <= r; i++)        a[i] = tmp[i];}void MergeSort( int l, int r){    if( l < r)    {        int mid = (l+r)>>1;        MergeSort(l,mid);        MergeSort(mid+1,r);        Merge(l,mid,r);    }}int main(){    while( ~scanf("%d%lld",&n,&k))    {        for( int i = 0; i < n; i++)            scanf("%d",&a[i]);        ans = 0;        MergeSort(0,n-1);        printf("%lld\n",max(0LL,ans-k));    }    return 0;}