Codeforces Round #295 Div1 B（Cubes）

Problem

Limits

$Time Limit(ms): 3000$

$Memory Limit(MB): 256$

$M\in[1,10^5]$

$X_i\in[-10^9,10^9]$

$Y_i\in[0,10^9]$

Look up Original Problem From here

Solution

一个点可取，当且仅当，把它取了之后，上面的点不会失去平衡而掉下来。

开两个优先队列$q1,q2$。$q1$的顶元素最大，$q2$的顶元素最小，起初把所有可取的点都放入$q1,q2$，然后，轮流从$q1,q2$取点，如果访问过了就取下一个，取出点后，判断这个点是否可取，如果不可取则取下一个…每次取出的点，判断$(X_i-1,Y_i-1),(X_i,Y_i-1),(X_i+1,Y_i-1)$这三个点是否可取，如果可取，则加入$q1,q2$。这样就可以得到M进制数。

Complexity

$Time Complexity: O(M\times log_2M)$

$Memory Complexity: O(M)$

My Code

//Hello. I‘m Peter.
#include<cstdio>
#include<iostream>
#include<sstream>
#include<cstring>
#include<string>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<cctype>
#include<ctime>
#include<stack>
#include<queue>
#include<vector>
#include<set>
#include<map>
using namespace std;
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
typedef unsigned int uin;
#define peter cout<<"i am peter"<<endl
#define input freopen("data.txt","r",stdin)
#define randin srand((unsigned int)time(NULL))
#define INT (0x3f3f3f3f)*2
#define LL (0x3f3f3f3f3f3f3f3f)*2
#define gsize(a) (int)a.size()
#define len(a) (int)strlen(a)
#define slen(s) (int)s.length()
#define pb(a) push_back(a)
#define clr(a) memset(a,0,sizeof(a))
#define clr_minus1(a) memset(a,-1,sizeof(a))
#define clr_INT(a) memset(a,INT,sizeof(a))
#define clr_true(a) memset(a,true,sizeof(a))
#define clr_false(a) memset(a,false,sizeof(a))
#define clr_queue(q) while(!q.empty()) q.pop()
#define clr_stack(s) while(!s.empty()) s.pop()
#define rep(i, a, b) for (int i = a; i < b; i++)
#define dep(i, a, b) for (int i = a; i > b; i--)
#define repin(i, a, b) for (int i = a; i <= b; i++)
#define depin(i, a, b) for (int i = a; i >= b; i--)
#define pi 3.1415926535898
#define eps 1e-9
#define MOD 1000000007
#define MAXN
#define N 100100
#define M
priority_queue<int>qbig;
priority_queue<int, vector<int>, greater<int> >qsmall;
int m;
bool vis[N];
map<pair<int,int>,int>mapit;
struct Point{
    int x,y;
}poi[N];
int dx[6]={-1,0,1,-1,0,1};
int dy[6]={-1,-1,-1,1,1,1};
const ll mod=1e9 + 9;
bool can_move(Point p0){
    rep(i,3,6){
        Point p1;
        p1.x=p0.x+dx[i];
        p1.y=p0.y+dy[i];
        pair<int,int>p=make_pair(p1.x,p1.y);
        if(mapit.find(p)!=mapit.end()){
            int t1=mapit[p];
            if(vis[t1]) continue;
            bool ok=false;
            rep(j,0,3){
                Point p2;
                p2.x=p1.x+dx[j];
                p2.y=p1.y+dy[j];
                if(p2.x==p0.x && p2.y==p0.y) continue;
                pair<int,int>p=make_pair(p2.x,p2.y);
                if(mapit.find(p)!=mapit.end()){
                    int t1=mapit[p];
                    if(vis[t1]) continue;
                    ok=true;
                    break;
                }
            }
            if(!ok) return false;
        }
    }
    return true;
}
int main(){
    scanf("%d",&m);
    rep(i,0,m){
        int x,y;
        scanf("%d %d",&x,&y);
        pair<int,int>p=make_pair(x,y);
        mapit[p]=i;
        poi[i].x=x,poi[i].y=y;
        vis[i]=false;
    }
    rep(i,0,m){
        if(can_move(poi[i])){
            qbig.push(i);
            qsmall.push(i);
        }
    }
    int turn=-1;
    vector<ll>res;
    res.clear();
    while(1){
        int now=0;
        turn=(turn+1)%2;
        if(turn==0){//Vasya big
            while(!qbig.empty()){
                now=qbig.top();
                if(vis[now]){
                    qbig.pop();
                    continue;
                }
                else if(!can_move(poi[now])){
                    qbig.pop();
                    continue;
                }
                else break;
            }
            if(qbig.empty()) break;
            qbig.pop();
            vis[now]=true;
            res.pb(now);
            pair<int,int>p=make_pair(poi[now].x,poi[now].y);
            mapit.erase(p);
            rep(i,0,3){
                int x=poi[now].x+dx[i];
                int y=poi[now].y+dy[i];
                pair<int,int>p=make_pair(x,y);
                if(mapit.find(p)!=mapit.end()){
                    int t1=mapit[p];
                    if(vis[t1]) continue;
                    if(can_move(poi[t1])){
                        qbig.push(t1);
                        qsmall.push(t1);
                    }
                }
            }
        }
        else if(turn==1){//.. small
            while(!qsmall.empty()){
                now=qsmall.top();
                if(vis[now]){
                    qsmall.pop();
                    continue;
                }
                else if(!can_move(poi[now])){
                    qsmall.pop();
                    continue;
                }
                else break;
            }
            if(qsmall.empty()) break;
            qsmall.pop();
            vis[now]=true;
            res.pb(now);
            pair<int,int>p=make_pair(poi[now].x,poi[now].y);
            mapit.erase(p);
            rep(i,0,3){
                int x=poi[now].x+dx[i];
                int y=poi[now].y+dy[i];
                pair<int,int>p=make_pair(x,y);
                if(mapit.find(p)!=mapit.end()){
                    int t1=mapit[p];
                    if(vis[t1]) continue;
                    if(can_move(poi[t1])){
                        qbig.push(t1);
                        qsmall.push(t1);
                    }
                }
            }
        }
    }
    int len=gsize(res);
    ll m1=1,ans=0;
    depin(i,len-1,0){
        ans=(ans+((res[i]*m1)%mod))%mod;
        m1=(m1*m)%mod;
    }
    printf("%lld\n",ans);
}