Inversion
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 2003 Accepted Submission(s): 787
Problem Description
bobo 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.
Input
The 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).
Output
For each tests:
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
Author
Xiaoxu Guo (ftiasch)
Source
2014 Multi-University Training Contest 5
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题意：求n个数的逆序对数，可以交换k次相邻的，所以求出原序列的逆序对后减去k即可。
思路：求逆序对有两种方法，归并排序和树状数组。逆序对的几种求法
代码：
//树状数组
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define maxn 100005
#define MAXN 2005
#define mod 1000000009
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-6
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
#define FRE(i,a,b) for(i = a; i <= b; i++)
#define FREE(i,a,b) for(i = a; i >= b; i--)
#define FRL(i,a,b) for(i = a; i < b; i++)
#define FRLL(i,a,b) for(i = a; i > b; i--)
#define mem(t, v) memset ((t) , v, sizeof(t))
#define sf(n) scanf("%d", &n)
#define sff(a,b) scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf printf
#define DBG pf("Hi\n")
typedef long long ll;
using namespace std;
int n,k,len;
int bit[maxn];
int a[maxn],num[maxn];
map<int,int>pos;
int cmp(int a,int b)
{
return a>b;
}
int Lowbit(int i)
{
return i&-i;
}
ll sum(int i)
{
ll s=0;
while (i>0){
s+=bit[i];
i-=Lowbit(i);
}
return s;
}
void add(int i,int x)
{
while (i<maxn)
{
bit[i]+=x;
i+=Lowbit(i);
}
}
ll solve()
{
int i,j;
ll ans=0;
FRE(i,1,n)
{
ans+=sum(pos[num[i]]-1);
add(pos[num[i]],1);
}
ans-=k;
if (ans<0) ans=0;
return ans;
}
int main()
{
int i,j;
while (~sff(n,k))
{
pos.clear();
mem(bit,0);
FRE(i,1,n)
{
sf(a[i]);
num[i]=a[i];
}
sort(a+1,a+n+1,cmp);
len=1;
pos[a[1]]=len;
FRE(i,2,n) //map去重，离散化
if (a[i]!=a[i-1])
pos[a[i]]=++len;
pf("len=%d\n",len);
pf("%I64d\n",solve());
}
return 0;
}
//归并排序
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define maxn 100005
#define MAXN 2005
#define mod 1000000009
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-6
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
#define FRE(i,a,b) for(i = a; i <= b; i++)
#define FREE(i,a,b) for(i = a; i >= b; i--)
#define FRL(i,a,b) for(i = a; i < b; i++)
#define FRLL(i,a,b) for(i = a; i > b; i--)
#define mem(t, v) memset ((t) , v, sizeof(t))
#define sf(n) scanf("%d", &n)
#define sff(a,b) scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf printf
#define DBG pf("Hi\n")
typedef long long ll;
using namespace std;
int a[maxn];
int n,k;
__int64 ans;
void Merge(int a[],int l,int mid,int r)
{
int i,j,k=l;
int *b=new int[r+1];
FRE(i,l,r) b[i]=a[i];
i=l,j=mid+1;
while (i<=mid&&j<=r)
{
if (b[i]<=b[j])
a[k++]=b[i++];
else{
a[k++]=b[j++];
ans+=(mid-i+1);
}
}
while (i<=mid) a[k++]=b[i++];
while (j<=r) a[k++]=b[j++];
delete []b;
}
void Merge_sort(int a[],int l,int r)
{
if (l==r) return ;
int mid=(l+r)>>1;
Merge_sort(a,l,mid);
Merge_sort(a,mid+1,r);
Merge(a,l,mid,r);
}
int main()
{
int i,j;
while (~sff(n,k))
{
FRE(i,1,n)
sf(a[i]);
ans=0;
Merge_sort(a,1,n);
ans-=k;
if (ans<0) ans=0;
pf("%I64d\n",ans);
}
return 0;
}