Codeforces 260E

首先9！枚举儿子的位置，
1.画出水平的两根线，把平面分成三块，每块的个数是个固定值，
2.画出竖直的两根线，把平面分成三块，每块的个数是个固定值，
3.四根线合起来，计数
也不知道这个代码是复制谁的。。。

#include<iostream>#include<cstdio>#include<map>#include<cstring>#include<cmath>#include<vector>#include<algorithm>#include<set>#include<string>#include<queue>#define inf 1000000005#define M 40#define N 100005#define maxn 300005#define eps 1e-12#define zero(a) fabs(a)<eps#define Min(a,b) ((a)<(b)?(a):(b))#define Max(a,b) ((a)>(b)?(a):(b))#define pb(a) push_back(a)#define mp(a,b) make_pair(a,b)#define mem(a,b) memset(a,b,sizeof(a))#define LL long long#define MOD 1000000007#define lson step<<1#define rson step<<1|1#define sqr(a) ((a)*(a))#define Key_value ch[ch[root][1]][0]#define test puts("OK");#define pi acos(-1.0)#define lowbit(x) ((-(x))&(x))#define HASH1 1331#define HASH2 10001#pragma comment(linker, "/STACK:1024000000,1024000000")using namespace std;struct Set_tree{    int left,right;    vector<int>v;}L[N*4];struct Point{    int x,y;    bool operator<(const Point n)const{        return x!=n.x?x<n.x:y<n.y;    }}p[N];int n,x[N],y[N];int a[9],b[9];double ret_x1,ret_x2,ret_y1,ret_y2;void Bulid(int step,int l,int r){    L[step].left=l;    L[step].right=r;    L[step].v.clear();    for(int i=l;i<=r;i++)        L[step].v.pb(p[i].y);    sort(L[step].v.begin(),L[step].v.end());    if(l==r)        return;    int m=(l+r)>>1;    Bulid(lson,l,m);    Bulid(rson,m+1,r);}int Query(int step,int l,int r,int val){    if(L[step].left==l&&r==L[step].right){        if(L[step].v.size()==0) return 0;        if(L[step].v[0]>val) return 0;        if(L[step].v.back()<=val) return L[step].v.size();        return (upper_bound(L[step].v.begin(),L[step].v.end(),val)-L[step].v.begin());    }    int m=(L[step].left+L[step].right)>>1;    if(r<=m) return Query(lson,l,r,val);    else if(l>m) return Query(rson,l,r,val);    else return Query(lson,l,m,val)+Query(rson,m+1,r,val);}bool ok(){    int x1=b[a[0]]+b[a[1]]+b[a[2]]-1;    int x2=x1+b[a[3]]+b[a[4]]+b[a[5]];    int y1=b[a[0]]+b[a[3]]+b[a[6]]-1;    int y2=y1+b[a[1]]+b[a[4]]+b[a[7]];    if(x1+1>=n||x[x1]==x[x1+1]) return false;    if(x2+1>=n||x[x2]==x[x2+1]) return false;    if(y1+1>=n||y[y1]==y[y1+1]) return false;    if(y2+1>=n||y[y2]==y[y2+1]) return false;    if(Query(1,0,x1,y[y1])!=b[a[0]]) return false;    if(Query(1,0,x1,y[y2])!=b[a[0]]+b[a[1]]) return false;    if(Query(1,x1+1,x2,y[y1])!=b[a[3]]) return false;    if(Query(1,x1+1,x2,y[y2])!=b[a[3]]+b[a[4]]) return false;    ret_x1=(x[x1]+x[x1+1])/2.0;    ret_x2=(x[x2]+x[x2+1])/2.0;    ret_y1=(y[y1]+y[y1+1])/2.0;    ret_y2=(y[y2]+y[y2+1])/2.0;    return true;}int main(){    //freopen("input.txt","r",stdin);    while(scanf("%d",&n)!=EOF){        for(int i=0;i<n;i++){            scanf("%d%d",&p[i].x,&p[i].y);            x[i]=p[i].x;y[i]=p[i].y;        }        sort(p,p+n);        sort(x,x+n);        sort(y,y+n);        Bulid(1,0,n-1);        for(int i=0;i<9;i++) scanf("%d",&b[i]);        for(int i=0;i<9;i++) a[i]=i;        int t=362880;        while(t--){            if(ok()){                printf("%.1f %.1f\n%.1f %.1f\n",ret_x1,ret_x2,ret_y1,ret_y2);                break;            }            next_permutation(a,a+9);        }        if(t<=0) puts("-1");    }    return 0;}