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BZOJ2626: JZPFAR

题目:http://www.lydsy.com/JudgeOnline/problem.php?id=2626

题解:裸K-Dtree,最大值?自己yy一下估价函数就好了。 

        两题居然是同一个错误,真是too naive。。。

代码:

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  1 #include<cstdio>  2 #include<cstdlib>  3 #include<cmath>  4 #include<cstring>  5 #include<algorithm>  6 #include<iostream>  7 #include<vector>  8 #include<map>  9 #include<set> 10 #include<queue> 11 #include<string> 12 #define inf 1ll<<62 13 #define maxn 200000+5 14 #define maxm 100000+5 15 #define eps 1e-10 16 #define ll double 17 #define pa pair<ll,int> 18 #define for0(i,n) for(int i=0;i<=(n);i++) 19 #define for1(i,n) for(int i=1;i<=(n);i++) 20 #define for2(i,x,y) for(int i=(x);i<=(y);i++) 21 #define for3(i,x,y) for(int i=(x);i>=(y);i--) 22 #define for4(i,x) for(int i=head[x],y=e[i].go;i;i=e[i].next,y=e[i].go) 23 #define mod 1000000007 24 #define sqr(x) (x)*(x) 25 using namespace std; 26 inline int read() 27 { 28     int x=0,f=1;char ch=getchar(); 29     while(ch<0||ch>9){if(ch==-)f=-1;ch=getchar();} 30     while(ch>=0&&ch<=9){x=10*x+ch-0;ch=getchar();} 31     return x*f; 32 } 33 int n,m,cur; 34 priority_queue<pa,vector<pa>,greater<pa> >q; 35 struct rec 36 { 37     int mi[2],mx[2],d[2],l,r,id; 38     int& operator [](int i){return d[i];} 39 }p[maxn],t[maxn],now; 40 bool operator <(rec a,rec b){return a[cur]<b[cur];} 41 inline void pushup(int k) 42 { 43     int l=t[k].l,r=t[k].r; 44     for0(i,1) 45     { 46         t[k].mi[i]=min(t[k][i],min(t[l].mi[i],t[r].mi[i])); 47         t[k].mx[i]=max(t[k][i],max(t[l].mx[i],t[r].mx[i])); 48     } 49 } 50 inline int build(int l,int r,int dir) 51 { 52     int mid=(l+r)>>1; 53     cur=dir; 54     nth_element(p+l,p+mid,p+r+1); 55     t[mid]=p[mid]; 56     for0(i,1)t[mid].mi[i]=t[mid].mx[i]=t[mid][i]; 57     t[mid].l=l>mid-1?0:build(l,mid-1,dir^1); 58     t[mid].r=mid+1>r?0:build(mid+1,r,dir^1); 59     pushup(mid); 60     return mid; 61 } 62 inline ll calc(int k) 63 { 64     if(!k)return -inf-1; 65     ll ret=0; 66     for0(i,1)ret+=max((ll)sqr(now[i]-t[k].mi[i]),(ll)sqr(t[k].mx[i]-now[i])); 67     return ret; 68 } 69 inline ll dist(rec a,rec b){return (ll)sqr(a[0]-b[0])+(ll)sqr(a[1]-b[1]);} 70 inline void query(int k) 71 { 72     if(!k)return; 73     ll dl=calc(t[k].l),dr=calc(t[k].r),d=dist(t[k],now); 74     if(d>q.top().first||(d==q.top().first&&t[k].id<-q.top().second))q.pop(),q.push(pa(d,-t[k].id)); 75     if(dl>dr) 76     { 77         if(dl>=q.top().first)query(t[k].l); 78         if(dr>=q.top().first)query(t[k].r); 79     }else 80     { 81         if(dr>=q.top().first)query(t[k].r); 82         if(dl>=q.top().first)query(t[k].l); 83     } 84 } 85      86 int main() 87 { 88     freopen("input.txt","r",stdin); 89     freopen("output.txt","w",stdout); 90     n=read(); 91     for1(i,n)p[i][0]=read(),p[i][1]=read(),p[i].id=i; 92     for0(i,1)t[0].mi[i]=1000000000,t[0].mx[i]=-1000000000; 93     int rt=build(1,n,0); 94     m=read(); 95     while(m--) 96     { 97         now[0]=read();now[1]=read();int k=read(); 98         while(!q.empty())q.pop(); 99         for1(i,k)q.push(pa(-inf,0));100         query(rt);101         printf("%d\n",-q.top().second);102     }103     return 0;104 }
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BZOJ2626: JZPFAR