标签：i++ std problem func lin update return online upd
想法：将(A,B)当成二维坐标。然后再转换为(A-B,A+B)。问题变成最小化$\sum_{i=L}^{R} {|A_i-A_0|+|B_i-B_0|} $。每一维是独立的，并且每一维选中位数是最优的之一。所以用主席树维护一下查第K大，前\后缀和。
Code $O(n \log n)$
#include < cstdio >
#include < algorithm >
#define gec getchar
#define FILE(F) freopen(F".in","r",stdin),freopen(F".out","w",stdout)
#define DEBUG fprintf(stderr,"Passing [%s] in Line (%d)\n",__FUNCTION__,__LINE__)
typedef long long ll;
template < typename T >
inline void read(T &x)
{
x=0;bool f=0; char c=gec();
for(;c<‘0‘||c>‘9‘;c=gec())f=(c==‘-‘);
for(;c>=‘0‘&&c<=‘9‘;c=gec())x=x*10+c-‘0‘;
x=f?-x:x;
}
const int MAXN(100010);
int n,q,A[MAXN],B[MAXN],X[MAXN],Y[MAXN];
struct AXLE
{
int a[MAXN],up;
void ins(int x){a[++up]=x;}
void build()
{
std::sort(a+1,a+1+up);
int _up=1;
for(int i=2;i<=up;i++)if(a[i]!=a[i-1])a[++_up]=a[i];
up=_up;
}
int Find(int x)
{
int l=1,r=up,mid,id=1;
while(l<=r)if(a[mid=(l+r)>>1]<=x)l=mid+1,id=mid;else r=mid-1;
return id;
}
}axle;
namespace ChairMan_Tree
{
const int MAX_L(2000010);
struct CMT
{
int nx[MAX_L][2],Siz[MAX_L]; ll Sum[MAX_L];
int root[MAXN],stot,Val[MAXN],N;
void update(int k)
{Sum[k]=Sum[nx[k][0]]+Sum[nx[k][1]];
Siz[k]=Siz[nx[k][0]]+Siz[nx[k][1]];}
void modfiy(int&k,int k2,int L,int R,int x)//val[x]++;
{
if(!k)k=++stot,Sum[k]=Sum[k2],Siz[k]=Siz[k2];
if(L==R){Sum[k]+=Val[x];Siz[k]++; return ;}
int MID=(L+R)>>1;
if(x<=MID)modfiy(nx[k][0],nx[k2][0],L,MID,x),nx[k][1]=nx[k2][1];
else modfiy(nx[k][1],nx[k2][1],MID+1,R,x),nx[k][0]=nx[k2][0];
update(k);
}
ll Que(int k1,int k2,int Rank)//查询区间第K大以及前\后缀和
{
int L=1,R=N,MID,Siz_L=0,Siz_R=0;
ll Sum_L=0,Sum_R=0;
while(L!=R)
{
MID=(L+R)>>1;
// fprintf(stderr,"%d %d %d\n",L,R,Rank);
// fprintf(stderr,"%d\n",Siz[nx[k1][0]]-Siz[nx[k2][0]]);
if(Siz[nx[k1][0]]-Siz[nx[k2][0]]<Rank)
Siz_L+=Siz[nx[k1][0]]-Siz[nx[k2][0]],Sum_L+=Sum[nx[k1][0]]-Sum[nx[k2][0]],
// fprintf(stderr,"%d %d %lld\n",L,MID,Sum[nx[k1][0]]-Sum[nx[k2][0]]),
L=MID+1,Rank-=Siz[nx[k1][0]]-Siz[nx[k2][0]],k1=nx[k1][1],k2=nx[k2][1];
else
Siz_R+=Siz[nx[k1][1]]-Siz[nx[k2][1]],Sum_R+=Sum[nx[k1][1]]-Sum[nx[k2][1]],
R=MID,k1=nx[k1][0],k2=nx[k2][0];
}
// fprintf(stderr,"%lld %d %lld %d %d %d\n",Sum_L,Siz_L,Sum_R,Siz_R,L,Val[L]);
return (ll)Siz_L*Val[L]-Sum_L+Sum_R-(ll)Siz_R*Val[L];
}
}A_Tree,B_Tree;
}using namespace ChairMan_Tree;
void Pretreat()
{
for(int i=1;i<=n;i++)axle.ins(A[i]); axle.build();
for(int i=1;i<=n;i++)A[i]=axle.Find(A[i]);
A_Tree.N=axle.up; for(int i=1;i<=A_Tree.N;i++)A_Tree.Val[i]=axle.a[i];
for(int i=1;i<=n;i++)A_Tree.modfiy(A_Tree.root[i],A_Tree.root[i-1],1,A_Tree.N,A[i]);
axle.up=0;
for(int i=1;i<=n;i++)axle.ins(B[i]); axle.build();
for(int i=1;i<=n;i++)B[i]=axle.Find(B[i]);
B_Tree.N=axle.up; for(int i=1;i<=B_Tree.N;i++)B_Tree.Val[i]=axle.a[i];
for(int i=1;i<=n;i++)B_Tree.modfiy(B_Tree.root[i],B_Tree.root[i-1],1,B_Tree.N,B[i]);
}
int main()
{
#ifndef ONLINE_JUDGE
FILE("C");
#endif
read(n);read(q);
for(int i=1;i<=n;i++)read(X[i]);
for(int i=1;i<=n;i++)read(Y[i]);
for(int i=1;i<=n;i++)A[i]=X[i]-Y[i],B[i]=X[i]+Y[i];
Pretreat();
for(int i=1,L,R;i<=q;i++)
{
read(L);read(R);
ll Ans_A=A_Tree.Que(A_Tree.root[R],A_Tree.root[L-1],(R-L+2)>>1);
ll Ans_B=B_Tree.Que(B_Tree.root[R],B_Tree.root[L-1],(R-L+2)>>1);
printf("%.2lf\n",(Ans_A+Ans_B)*0.5);
}
return 0;
}
BZOJ 2735: 世博会 主席树+切比雪夫距离转曼哈顿距离
标签：i++ std problem func lin update return online upd
原文地址：http://www.cnblogs.com/Oncle-Ha/p/7123627.html
