Codeforces 520D. Cubes 贪心模拟

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每一步都取当前稳定的格子里面数字最大或者最小的数.

用一个set维护当前可取的格子 *begin 最大 *(--end) 最小

每删除一个格子都要对这个格子周围的6个格子进行稳定性检查

一个格子上面的3个格子确定了这个格子的稳定性

Once Vasya and Petya assembled a figure of m cubes, each of them is associated with a number between 0 and m - 1 (inclusive, each number appeared exactly once). Let's consider a coordinate system such that the OX is the ground, and the OY is directed upwards. Each cube is associated with the coordinates of its lower left corner, these coordinates are integers for each cube.

The figure turned out to be stable. This means that for any cube that is not on the ground, there is at least one cube under it such that those two cubes touch by a side or a corner. More formally, this means that for the cube with coordinates (x, y) either y = 0, or there is a cube with coordinates (x - 1, y - 1), (x, y - 1) or (x + 1, y - 1).

Now the boys want to disassemble the figure and put all the cubes in a row. In one step the cube is removed from the figure and being put to the right of the blocks that have already been laid. The guys remove the cubes in such order that the figure remains stable. To make the process more interesting, the guys decided to play the following game. The guys take out the cubes from the figure in turns. It is easy to see that after the figure is disassembled, the integers written on the cubes form a number, written in the m-ary positional numerical system (possibly, with a leading zero). Vasya wants the resulting number to be maximum possible, and Petya, on the contrary, tries to make it as small as possible. Vasya starts the game.

Your task is to determine what number is formed after the figure is disassembled, if the boys play optimally. Determine the remainder of the answer modulo 109 + 9.

Input

The first line contains number m (2 ≤ m ≤ 105).

The following m lines contain the coordinates of the cubes xi, yi ( - 109 ≤ xi ≤ 109, 0 ≤ yi ≤ 109) in ascending order of numbers written on them. It is guaranteed that the original figure is stable.

No two cubes occupy the same place.

Output

In the only line print the answer to the problem.

Sample test(s)
input
32 11 00 1
output
19
input
50 00 10 20 30 4
output
2930

[cpp] view plain copy
/* *********************************************** 
Author        :CKboss 
Created Time  :2015Äê03ÔÂ03ÈÕ ÐÇÆÚ¶þ 23Ê±46·Ö15Ãë 
File Name     :D.cpp 
************************************************ */  
  
#include <iostream>  
#include <cstdio>  
#include <cstring>  
#include <algorithm>  
#include <string>  
#include <cmath>  
#include <cstdlib>  
#include <vector>  
#include <queue>  
#include <set>  
#include <map>  
  
using namespace std;  
  
typedef long long int LL;  
  
const LL mod = 1e9+9;  
const int maxn=100100;  
  
struct BOX  
{  
    int x,y,id;  
}box[maxn];  
  
typedef pair<int,int> pII;  
  
int n;  
map<pII,int> mPI;  
set<pII> Pt;  
set<int> St;  
  
bool isST(int id)  
{  
    int x=box[id].x;  
    int y=box[id].y;  
  
    pII tp;  
  
    /// check (x-1,y+1)  
    tp = make_pair(x-1,y+1);  
    if(Pt.count(tp)==1)  
    {  
        if(Pt.count( make_pair(x-2,y) )   
                || Pt.count( make_pair(x-1,y) ))  
            ;  
        else return false;  
    }  
  
    /// check (x,y+1)  
    tp = make_pair(x,y+1);  
    if(Pt.count(tp)==1)  
    {  
        if(Pt.count( make_pair(x-1,y) )  
                    || Pt.count( make_pair(x+1,y) ))  
            ;  
        else return false;  
    }  
  
    /// check (x+1,y+1)  
    tp = make_pair(x+1,y+1);  
    if(Pt.count(tp)==1)  
    {  
        if(Pt.count( make_pair(x+1,y) )  
                    || Pt.count( make_pair(x+2,y) ))  
            ;  
        else return false;  
    }  
  
    return true;  
}  
  
void checkST(int id)  
{  
    if(isST(id))  
    {  
        St.insert(id);  
    }  
    else  
    {  
        if(St.count(id)) St.erase(id);  
    }  
}  
  
void checkround(int id)  
{  
    int x=box[id].x;  
    int y=box[id].y;  
      
    checkST(id);  
      
    for(int i=-1;i<=1;i++) /// dx  
    {  
        for(int j=-1;j<=1;j+=2) /// dy  
        {  
            int nx=x+i,ny=y+j;  
            pII tp = make_pair(nx,ny);  
            if(Pt.count(tp)==1)  
            {  
                int dd=mPI[tp];  
                checkST(dd);  
            }  
        }  
    }  
}  
  
void Remove(int id)  
{  
    int x=box[id].x;  
    int y=box[id].y;  
  
    /// erase  
    Pt.erase(make_pair(x,y));  
    St.erase(id);  
  
    /// check other stable box  
      
    for(int i=-1;i<=1;i++) // dx  
    {  
        for(int j=-1;j<=1;j+=2) // dy  
        {  
            int nx=x+i,ny=y+j;  
            pII tp = make_pair(nx,ny);  
            if(Pt.count(tp)==1)  
            {  
                int ID=mPI[tp];  
                checkround(ID);  
            }  
        }  
    }  
      
}  
  
int main()  
{  
    //freopen("in.txt","r",stdin);  
    //freopen("out.txt","w",stdout);  
  
    scanf("%d",&n);  
    for(int i=0;i<n;i++)  
    {  
        int x,y;  
        scanf("%d%d",&x,&y);  
        box[i]=(BOX){x,y,i};  
        mPI[make_pair(x,y)]=i;  
        Pt.insert(make_pair(x,y));  
    }  
    /// check stable disassemble box  
  
    int mxid=0,miid=0;  
    for(int i=0;i<n;i++)  
    {  
        if(isST(i))  
        {  
            St.insert(i);  
            mxid=max(mxid,i);  
            miid=min(miid,i);  
        }  
    }  
  
    vector<int> vi;  
      
    for(int loop=0;loop<n;loop++)  
    {  
        if(loop%2==0) //find max  
        {  
            vi.push_back(mxid);  
            Remove(mxid);  
        }  
        else // find min  
        {  
            vi.push_back(miid);  
            Remove(miid);  
        }  
        if(loop==n-1) continue;  
        miid=*St.begin();  
        mxid=*(--St.end());  
    }  
  
    LL ans=0,base=1;  
    for(int sz=vi.size(),i=sz-1;i>=0;i--)  
    {  
        ans=(ans+base*vi[i]%mod)%mod;  
        base=(base*n)%mod;  
    }  
  
    cout<<ans%mod<<endl;  
      
    return 0;  
}  