Codeforces Round #295 D. Cubes [贪心 set map]

传送门

D. Cubes

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 x_i, y_i ( - 10⁹ ≤ x_i ≤ 10⁹, 0 ≤ y_i ≤ 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

题意：有m个方块 每个方块有一个值 并且是堆起来稳定的 一个方块可以拿掉当且仅当剩下的还是稳定的 双方轮流拿 从左到右放组成一个m进制的数 （转自 http://blog.csdn.net/u011686226/article/details/44036875）

10129550	2015-03-03 11:00:30	njczy2010	D - Cubes	GNU C++	Accepted	499 ms	32596 KB
10129479	2015-03-03 10:52:18	njczy2010	D - Cubes	GNU C++	Wrong answer on test 21	421 ms	21600 KB
10129457	2015-03-03 10:49:33	njczy2010	D - Cubes	GNU C++	Wrong answer on test 21	436 ms	21700 KB
10129412	2015-03-03 10:44:12	njczy2010	D - Cubes	GNU C++	Wrong answer on test 3	0 ms	41100 KB

 #include<iostream>
 #include<cstring>
 #include<cstdlib>
 #include<cstdio>
 #include<algorithm>
 #include<cmath>
 #include<queue>
 #include<map>
 #include<set>
 #include<stack>
 #include<string>

 #define N 100005
 #define M 10005
 #define mod 1000000007
 //#define p 10000007
 //#define mod2 1000000009
 #define ll long long
 #define ull unsigned long long
 #define LL long long
 #define eps 1e-6
 //#define inf 2147483647
 #define maxi(a,b) (a)>(b)? (a) : (b)
 #define mini(a,b) (a)<(b)? (a) : (b)

 using namespace std;

 int n;
 int x[N],y[N];
 map<pair<int,int>,int>mp;
 int r[N];
 set<int>s;
 int ans[N];
 int vis[N];
 ll mod2=;

 int ok(int i)
 {
     int nx,ny,te;
     nx=x[i]-;ny=y[i]+;
     te=mp[ make_pair(nx,ny) ];
     if(te!= && r[te-]==){
         return ;
     }
     nx=x[i];
     te=mp[ make_pair(nx,ny) ];
     if(te!= && r[te-]==){
         return ;
     }
     nx=x[i]+;
     te=mp[ make_pair(nx,ny) ];
     if(te!= && r[te-]==){
         return ;
     }
     return ;
 }

 void ini()
 {
     int i;
     int nx,ny;
     s.clear();
     memset(r,,sizeof(r));
     memset(vis,,sizeof(vis));
     for(i=;i<n;i++){
         scanf("%d%d",&x[i],&y[i]);
         mp[ make_pair(x[i],y[i]) ]=i+;
     }
     for(i=;i<n;i++){
         nx=x[i]-;ny=y[i]-;
         if(mp[ make_pair(nx,ny) ]!=){
             r[i]++;
         }
         nx=x[i];
         if(mp[ make_pair(nx,ny) ]!=){
             r[i]++;
         }
         nx=x[i]+;
         if(mp[ make_pair(nx,ny) ]!=){
             r[i]++;
         }
     }
     for(i=;i<n;i++){
         if(ok(i)==){
             s.insert(i);
             vis[i]=-;
         }
     }
 }

 void changeerase(int i)
 {
     int nx,ny,te;
     nx=x[i]-;ny=y[i]-;
     te=mp[ make_pair(nx,ny) ];
     if(te!=){
         if(vis[te-]==-){
             vis[te-]=;s.erase(te-);
         }
         return;
     }
     nx=x[i];
     te=mp[ make_pair(nx,ny) ];
     if(te!=){
         if(vis[te-]==-){
             vis[te-]=;s.erase(te-);
         }
         return;
     }
     nx=x[i]+;
     te=mp[ make_pair(nx,ny) ];
     if(te!=){
         if(vis[te-]==-){
             vis[te-]=;s.erase(te-);
         }
         return;
     }
 }

 void changeadd(int i)
 {
     int nx,ny,te;
     nx=x[i]-;ny=y[i]-;
     te=mp[ make_pair(nx,ny) ];
     if(te!= && ok(te-)==){
         s.insert(te-);
         vis[te-]=-;
     }
     nx=x[i];
     te=mp[ make_pair(nx,ny) ];
     if(te!= && ok(te-)==){
         s.insert(te-);
         vis[te-]=-;
     }
     nx=x[i]+;
     te=mp[ make_pair(nx,ny) ];
     if(te!= && ok(te-)==){
         s.insert(te-);
         vis[te-]=-;
     }
 }

 void updata(int i)
 {
     int nx,ny,te;
     nx=x[i]-;ny=y[i]+;
     te=mp[ make_pair(nx,ny) ];
     if(te!=){
         r[te-]--;
         if(r[te-]==)
             changeerase(te-);
     }
     nx=x[i];
     te=mp[ make_pair(nx,ny) ];
     if(te!=){
         r[te-]--;
         if(r[te-]==)
             changeerase(te-);
     }
     nx=x[i]+;
     te=mp[ make_pair(nx,ny) ];
     if(te!=){
         r[te-]--;
         if(r[te-]==)
             changeerase(te-);
     }
 }

 void solve()
 {
     int i,index;
     set<int>::iterator it;
     for(i=;i<n;i++){
         if(i%==){
             it=s.end();
             it--;
             ans[i]=*it;
         }
         else{
             it=s.begin();
             ans[i]=*it;
         }
         index=*it;
         s.erase(*it);
         mp[ make_pair(x[index],y[index]) ]=;
         vis[index]=;
         updata(index);
         changeadd(index);
     }
 }

 void out()
 {
     int i;
     ll aa=;
     /*
     for(i=0;i<n;i++){
         printf("%d",ans[i]);
     }
     printf("\n");*/
     for(i=;i<n;i++){
         aa=(aa*(ll)n)%mod2;
         aa=(aa+(ll)ans[i])%mod2;
     }
     printf("%I64d\n",aa);
 }

 int main()
 {
     //freopen("data.in","r",stdin);
     //freopen("data.out","w",stdout);
     //scanf("%d",&T);
     //for(int ccnt=1;ccnt<=T;ccnt++)
     //while(T--)
     //scanf("%d%d",&n,&m);
     while(scanf("%d",&n)!=EOF)
     {
         ini();
         solve();
         out();
     }
     return ;
 }