HDU 5497 Inversion

 1 /*by SilverN*/
 2 #include<iostream>
 3 #include<algorithm>
 4 #include<cstring>
 5 #include<cstdio>
 6 #include<cmath>
 7 using namespace std;
 8 const int mxn=110000;
 9 int n,m;
10 struct tree{
11     int a[mxn];
12     inline lowbit(int x){return x&-x;}
13     void init(){memset(a,0,sizeof a);}
14     void add(int x,int num){
15         while(x<=n){
16             a[x]+=num;
17             x+=lowbit(x);    
18         }
19         return;
20     }
21     int sum(int x){
22         int res=0;
23         while(x){
24             res+=a[x];
25             x-=lowbit(x);
26         }
27         return res;
28     }
29 };
30 int a[mxn];
31 tree Le,Ri;
32 int main(){
33     int T;
34     scanf("%d",&T);
35     while(T--){
36         scanf("%d%d",&n,&m);
37         int i,j;
38         for(i=1;i<=n;i++){
39             scanf("%d",&a[i]);
40         }
41         Le.init();
42         Ri.init();
43         long long ans;
44         long long tmp=0;
45         //求原有逆序对数(最左边滑动窗口内的不算) 
46         for(i=n;i>m;i--){
47             Ri.add(a[i],1);
48             tmp+=Ri.sum(a[i]-1);
49         }
50         ans=tmp;
51         //finish
52         for(i=1;i<=n-m;i++){
53             Ri.add(a[i+m],-1);//窗口滑动，窗口最右面的数被删除
54             tmp=tmp+Ri.sum(a[i]-1)-Ri.sum(a[i+m]-1);
55             tmp=tmp+Le.sum(n+1-(a[i]+1))-Le.sum(n+1-(a[i+m]+1));
56             Le.add(n+1-a[i],1);
57             ans=min(ans,tmp);
58         }
59         printf("%lld\n",ans);
60     }
61     return 0;
62 }