【枚举】【权值分块】bzoj1112 [POI2008]砖块Klo

 1 #include<cstdio>
 2 #include<algorithm>
 3 #include<cmath>
 4 using namespace std;
 5 typedef long long ll;
 6 struct Point{int v,p;}t[100001];
 7 bool operator < (const Point &a,const Point &b){return a.v<b.v;}
 8 int sumv[350],ma[100001],en,a[100001],b[100001],tot[350],l[350],r[350],num[100001],sum=1,K,n,m;
 9 ll ans=999999999999999999,tAns;
10 void makeblock()
11 {
12     int sz=sqrt(en); if(!sz) sz=1;
13     for(;sum*sz<en;++sum)
14       {
15           l[sum]=r[sum-1]+1; r[sum]=sum*sz;
16           for(int i=l[sum];i<=r[sum];++i) num[i]=sum;
17       }
18     l[sum]=r[sum-1]+1; r[sum]=en;
19     for(int i=l[sum];i<=r[sum];++i) num[i]=sum;
20 }
21 void Insert(const int &x){++b[x]; ++tot[num[x]]; sumv[num[x]]+=(ll)ma[x];}
22 void Delete(const int &x){--b[x]; --tot[num[x]]; sumv[num[x]]-=(ll)ma[x];}
23 int Query(const int &x)
24 {
25     int cnt=0; ll now=0;
26     for(int i=1;;i++)
27       {
28         cnt+=tot[i]; now+=sumv[i];
29         if(cnt>=x)
30           {
31               cnt-=tot[i]; now-=sumv[i];
32             for(int j=l[i];;j++)
33               {
34                 cnt+=b[j]; now+=(ll)ma[j]*(ll)b[j];
35                 if(cnt>=x)
36                   {
37                       cnt-=b[j]; now-=(ll)ma[j]*(ll)b[j];
38                       tAns=((ll)ma[j]*(ll)cnt-now);
39                       return j;
40                   }
41               }
42           }
43       }
44 }
45 void Next_Sum(const int &x)
46 {
47     int cnt=0; ll now=0;
48     for(int i=x+1;i<=r[num[x]];++i) {cnt+=b[i]; now+=(ll)ma[i]*(ll)b[i];}
49     for(int i=num[x]+1;i<=sum;++i) {cnt+=tot[i]; now+=sumv[i];}
50     tAns+=(now-(ll)ma[x]*(ll)cnt);
51 }
52 int main()
53 {
54     scanf("%d%d",&n,&m); K=(m>>1)+1;
55     for(int i=1;i<=n;++i)
56       {
57           scanf("%d",&t[i].v);
58           t[i].p=i;
59       } sort(t+1,t+n+1);
60     ma[a[t[1].p]=++en]=t[1].v;
61     for(int i=2;i<=n;++i)
62       {
63           if(t[i].v!=t[i-1].v) ++en;
64           ma[a[t[i].p]=en]=t[i].v;
65       } makeblock();
66     for(int i=1;i<=m;++i) Insert(a[i]);
67     int t=Query(K); Next_Sum(t); ans=min(ans,tAns);
68     for(int i=m+1;i<=n;++i)
69       {
70           Delete(a[i-m]); Insert(a[i]);
71           int t=Query(K);
72         Next_Sum(t);
73         ans=min(ans,tAns);
74       } printf("%lld\n",ans);
75     return 0;
76 }