Codeforces Round #295 (Div. 1) B. Cubes

两个人轮流取方块，取完方块必须使剩下的稳定，先手要求取数大，后手反之。问最后的m进制数是多少。。。。用STL维护一下。。。

#include <iostream>
#include <queue> 
#include <stack> 
#include <map> 
#include <set> 
#include <bitset> 
#include <cstdio> 
#include <algorithm> 
#include <cstring> 
#include <climits>
#include <cstdlib>
#include <cmath>
#include <time.h>
#define maxn 100005
#define maxm 100005
#define eps 1e-7
#define mod 1000000009
#define INF 0x3f3f3f3f
#define PI (acos(-1.0))
#define lowbit(x) (x&(-x))
#define mp make_pair
#define ls o<<1
#define rs o<<1 | 1
#define lson o<<1, L, mid 
#define rson o<<1 | 1, mid+1, R
#define pii pair<int, int>
#pragma comment(linker, "/STACK:16777216")
typedef long long LL;
typedef unsigned long long ULL;
//typedef int LL;
using namespace std;
LL qpow(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base;base=base*base;b/=2;}return res;}
LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;}
// head

map<pair<int, int>, int> mpp;
set<pair<int, int> > s;
set<int> leaf;
int x[maxn];
int y[maxn];
int n;

void read()
{
	scanf("%d", &n);
	for(int i = 0; i < n; i++) {
		scanf("%d%d", &x[i], &y[i]);
		pair<int, int> t = mp(x[i], y[i]);
		mpp[t] = i;
		s.insert(t);
	}
}

int cnt(int xx, int yy)
{
	int res = 0;
	for(int i = xx - 1; i <= xx + 1; i++) if(s.find(mp(i, yy - 1)) != s.end()) res++;
	return res;
}

bool isleaf(int xx, int yy)
{
	for(int i = xx - 1; i <= xx + 1; i++)
		if(s.find(mp(i, yy + 1)) != s.end() && cnt(i, yy + 1) == 1) return false;
	return true;
}

void update(int xx, int yy)
{
	for(int i = xx - 2; i <= xx + 2; i++)
		for(int j = yy - 1; j <= yy + 1; j++) {
			pair<int, int> t = mp(i, j);
			if(s.find(t) == s.end()) continue;
			int id = mpp[t];
			if(leaf.find(id) != leaf.end()) leaf.erase(id);
			if(isleaf(i, j)) leaf.insert(id);
		}
}

void work()
{
	for(int i = 0; i < n; i++)
		if(isleaf(x[i], y[i])) leaf.insert(i);
	LL ans = 0;
	for(int i = 0; i < n; i++) {
		int id;
		if(i % 2 == 0) id = *(--leaf.end());
		else id = *(leaf.begin());
		ans = (ans * n + id) % mod;
		s.erase(mp(x[id], y[id]));
		leaf.erase(id);
		update(x[id], y[id]);
	}
	printf("%I64d\n", ans);
}

int main()
{
	read();
	work();

	return 0;
}