HDU 4911
Description
bobo has a sequence a 1,a 2,…,a n. 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 a i>a j.
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 a i>a j.
Input
The input consists of several tests. For each tests:
The first line contains 2 integers n,k (1≤n≤10 5,0≤k≤10 9). The second line contains n integers a 1,a 2,…,a n (0≤a i≤10 9).
The first line contains 2 integers n,k (1≤n≤10 5,0≤k≤10 9). The second line contains n integers a 1,a 2,…,a n (0≤a i≤10 9).
Output
For each tests:
A single integer denotes the minimum number of inversions.
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
题意:相邻的最多调换k次,使得逆对数最小,
思路,先求出整个序列的逆对数-k次就可,如果出现负数就输出为0.用归并排序求逆对数。值得注意的是千万不要使用暴力会超时,还有就要使用long long型不能也不行。输出也是一个值得注意地方。
程序代码:
#include<iostream> #include<cstring> #include<cstdio> #include<algorithm> #define M 100007 #define ll _int64 ll a[M],c[M],ans,k; using namespace std; void merg(ll a[],int first,int mid,int last,ll c[]) { int i=first,j=mid+1; int m=mid,n=last; 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++]; ans+=(m-i+1); } } for(i=0;i<k;i++) a[first+i]=c[i]; } void merge_sort(ll a[], int first,int last,ll c[]) { if(first<last) { int mid=(first+last)>>1; merge_sort(a,first,mid,c); merge_sort(a,mid+1,last,c); merg(a,first,mid,last,c); } } int main() { int n; ll k; while(scanf("%d%I64d",&n,&k)!=EOF) { memset(a,0,sizeof(a)); memset(c,0,sizeof(c)); ans=0; for(int i=0;i<n;i++) scanf("%I64d",&a[i]); merge_sort(a,0,n-1,c); if(ans-k<=0) printf("0\n"); else printf("%I64d\n",ans-k); } return 0; }