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Codeforces 520D. Cubes 貪心模擬

來源：程序員人生發布時間：2015-04-11 09:25:06 閱讀次數：3971次

每步都取當前穩定的格子里面數字最大或最小的數.

用1個set保護當前可取的格子 *begin 最大 *(--end) 最小

每刪除1個格子都要對這個格子周圍的6個格子進行穩定性檢查

1個格子上面的3個格子肯定了這個格子的穩定性

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
3
2 1
1 0
0 1

output
19

input
5
0 0
0 1
0 2
0 3
0 4

output
2930

/* *********************************************** 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 (x⑴,y+1) tp = make_pair(x⑴,y+1); if(Pt.count(tp)==1) { if(Pt.count( make_pair(x⑵,y) ) || Pt.count( make_pair(x⑴,y) )) ; else return false; } /// check (x,y+1) tp = make_pair(x,y+1); if(Pt.count(tp)==1) { if(Pt.count( make_pair(x⑴,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=⑴;i<=1;i++) /// dx { for(int j=⑴;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=⑴;i<=1;i++) // dx { for(int j=⑴;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==n⑴) continue; miid=*St.begin(); mxid=*(--St.end()); } LL ans=0,base=1; for(int sz=vi.size(),i=sz⑴;i>=0;i--) { ans=(ans+base*vi[i]%mod)%mod; base=(base*n)%mod; } cout<<ans%mod<<endl; return 0; }

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Codeforces 520D. Cubes 貪心模擬