D. Guess Your Way Out! II
time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Amr bought a new video game "Guess Your Way Out! II". The goal of the game is to find an exit from the maze that looks like a perfect binary tree of height h.
The player is initially standing at the root of the tree and the exit from the tree is located at some leaf node.

Let's index all the nodes of the tree such that

  • The root is number 1
  • Each internal node i (i ≤ 2h - 1 - 1)
    will have a left child with index = 2i and a right child with index = 2i + 1

The level of a node is defined as 1 for a root, or 1 +
level of parent of the node otherwise. The vertices of the level h are called leaves. The exit to the maze is located at some leaf node n,
the player doesn't know where the exit is so he has to guess his way out!

In the new version of the game the player is allowed to ask questions on the format "Does the ancestor(exit, i) node
number belong to the range [L, R]?". Here ancestor(v, i) is
the ancestor of a node v that located in the level i.
The game will answer with "Yes" or "No" only. The game is designed such that it doesn't always answer correctly, and sometimes it cheats to confuse the player!.

Amr asked a lot of questions and got confused by all these answers, so he asked you to help him. Given the questions and its answers, can you identify whether the game is telling contradictory information or not? If the information is not contradictory and
the exit node can be determined uniquely, output its number. If the information is not contradictory, but the exit node isn't defined uniquely, output that the number of questions is not sufficient. Otherwise output that the information is contradictory.

Input

The first line contains two integers h, q (1 ≤ h ≤ 50, 0 ≤ q ≤ 105),
the height of the tree and the number of questions respectively.

The next q lines will contain four integers each i, L, R, ans (1 ≤ i ≤ h, 2i - 1 ≤ L ≤ R ≤ 2i - 1, ),
representing a question as described in the statement with its answer (ans = 1 if the answer is "Yes" and ans = 0 if
the answer is "No").

Output

If the information provided by the game is contradictory output "Game cheated!" without the quotes.

Else if you can uniquely identify the exit to the maze output its index.

Otherwise output "Data not sufficient!" without the quotes.

Sample test(s)
input
3 1
3 4 6 0
output
7
input
4 3
4 10 14 1
3 6 6 0
2 3 3 1
output
14
input
4 2
3 4 6 1
4 12 15 1
output
Data not sufficient!
input
4 2
3 4 5 1
2 3 3 1
output
Game cheated!
Note

Node u is an ancestor of node v if
and only if

  • u is the same node as v,
  • u is the parent of node v,
  • or u is an ancestor of the parent of node v.

In the first sample test there are 4 leaf nodes 4, 5, 6, 7.
The first question says that the node isn't in the range [4, 6] so the exit is node number 7.

In the second sample test there are 8 leaf nodes. After the first question the exit is in the range [10, 14].
After the second and the third questions only node number 14 is correct. Check the picture below to fully understand.

有一堆区间,1个入口

给出例如以下条件,区间[L,R]有/无入口

问入口在哪?

用set维护

记得lower_bound(a) 是第一个>=a的

upper_bound(a) 是第一个>a的

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
#include<stack>
#include<set>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define MAXH (50+10)
#define MAXQ (100000+10)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
typedef long long ll;
typedef pair<ll,ll> pll;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return (a-b+(a-b)/F*F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
char s1[]="Game cheated!\n",s2[]="Data not sufficient!\n";
int h,q;
pll cro(pll p,ll l,ll r)
{
ll a=p.first,b=p.second;
if (b<l||r<a) return mp(-1,-1);
return mp(max(a,l),min(b,r));
}
set<pll > S;
stack<pll > ask;
int main()
{
// freopen("D.in","r",stdin);
// freopen(".out","w",stdout); cin>>h>>q;
pll ans=mp(1LL<<(h-1),(1LL<<h) - 1); ll L=1LL<<(h-1),R=(1LL<<h) - 1; For(qcase,q)
{
int i,b;
ll l,r;
cin>>i>>l>>r>>b;
while (i<h) l<<=1,r=(r<<1)^1,++i;
if (b) ans=cro(ans,l,r);
else ask.push(mp(l,r));
}
if (ans.fi==-1) {
cout<<s1;
return 0;
}
S.insert(ans);
while (!ask.empty())
{
pll now=ask.top();
ask.pop(); set<pll>::iterator it,it2;
it=S.upper_bound(now);
if (it!=S.begin()) it--;
for(;it!=S.end();)
{
pll pit=*it; if (now.se<pit.fi) break;
if (cro(now,pit.fi,pit.se).fi==-1) {
it++;continue;
}
it2 = it;
it2++; if (pit.fi<now.fi) S.insert(mp(pit.fi,now.fi-1));
if (pit.se>now.se) S.insert(mp(now.se+1,pit.se));
S.erase(it); it=it2;
} } // cout<<ans.first<<' '<<ans.second<<endl; if (S.empty())
{
cout<<s1;
return 0;
}
if (S.size()==1)
{
ll p1=S.begin()->fi,p2=S.begin()->se; if (p1==p2) {
cout<<p1<<endl;
return 0;
}
} cout<<s2; return 0;
}

CF 558D(Guess Your Way Out! II-set解决区间问题)的更多相关文章

  1. 区间合并 --- Codeforces 558D : Gess Your Way Out ! II

    D. Guess Your Way Out! II Problem's Link: http://codeforces.com/problemset/problem/558/D Mean: 一棵满二叉 ...

  2. codeforces 558D Guess Your Way Out! II 规律

    题目链接 题意: 给出n和q 表示有一棵深度为n的全然二叉树.叶子节点中有恰好一个点是出口 主角从根往下走.但不知道出口在哪里,但主角会获得q个提示.  像这样标号 q个提示 格式: deep [l, ...

  3. CF R 635 div1 C Kaavi and Magic Spell 区间dp

    LINK:Kaavi and Magic Spell 一打CF才知道自己原来这么菜 这题完全没想到. 可以发现 如果dp f[i][j]表示前i个字符匹配T的前j个字符的方案数 此时转移变得异常麻烦 ...

  4. LightOJ - 1245 Harmonic Number (II) 求同值区间的和

    题目大意:对下列代码进行优化 long long H( int n ) {    long long res = 0;    for( int i = 1; i <= n; i++ )      ...

  5. jump-game i&&ii 能否跳出区间 贪心

    I: Given an array of non-negative integers, you are initially positioned at the first index of the a ...

  6. LightOJ 1089 - Points in Segments (II) 线段树区间修改+离散化

    http://www.lightoj.com/volume_showproblem.php?problem=1089 题意:给出许多区间,查询某个点所在的区间个数 思路:线段树,由于给出的是区间,查询 ...

  7. FZU Problem 2171 防守阵地 II (线段树区间更新模板题)

    http://acm.fzu.edu.cn/problem.php?pid=2171 成段增减,区间求和.add累加更新的次数. #include <iostream> #include ...

  8. CF卡技术详解——笔记

    知识太全面了,摘抄摘不完,还是粘过来加上注释和笔记吧. 重点以及断句用加粗,注释用红括号. 一.CF卡技术及规格 一.CF卡技术及规格 1.CF卡简史 随着数码产品的高速普及,近年来闪存卡也进入了高速 ...

  9. CF A.Mishka and Contest【双指针/模拟】

    [链接]:CF/4892 [题意]: 一个人解决n个问题,这个问题的值比k小, 每次只能解决最左边的或者最右边的问题 解决了就消失了.问这个人能解决多少个问题. [代码]: #include<b ...

随机推荐

  1. POJ 2409

    水题一道,不加优化也能0MS #include <iostream> #include <cstdio> #include <algorithm> #include ...

  2. Spring boot 使用@Value注入属性

    Spring boot 使用@Value注入属性 学习了:http://blog.csdn.net/hry2015/article/details/72353994 如果启动的时候报错: spring ...

  3. SQL编码中注意的性能问题

    1.选择合适的数据类型 为列选择最小化的数据类型 假设一列中的文本长度不一,使用VARCHAR而不是CHAR 不存储Unicode不要使用NVARCHAR或者NCHAR 假设一行的长度不超过8000, ...

  4. nodejs即时聊天

    一直想做一个即时聊天的应用,前几天看到了socket.io,感觉还不错.自己略加改动,感觉挺不错的.官网上给的样例非常easy,以下改进了一点,实现了历史消息的推送. demo地址:chat.code ...

  5. Appium AndroidKeyCode

    driver.sendKeyEvent(AndroidKeyCode.BACK); driver.sendKeyEvent(AndroidKeyCode.BACKSPACE); driver.send ...

  6. MyEclipse中安装Testng插件

    下载testng.eclipse插件 Testng相应jar包,这里使用的是5.12的版本号 直接找到myeclipse的安装文件夹,将org.testng.eclipse_5.12.0.6.jar ...

  7. Testbench的编写

    Testbench的作用,在于给我们编写的可综合代码的模块送入激励.即在我们波形仿真中用编写testbench来代替拖拽波形.其中还包括了我们硬件仿真与matlab仿真的联调建立(将matlab产生的 ...

  8. Synergy 共享键盘和鼠标

    直接安装Synergy 不行的话加配置文件 ➜ ~ cat synergy.conf section: screens lab712-PC: ckboss-HP: end section: links ...

  9. BZOJ 4260 trie树

    思路: 搞一个前缀异或和 一次从左往右 另一次从右往左 异或最大值 用字典树搞一搞 //By SiriusRen #include <cstdio> #include <cstrin ...

  10. Windows窗体应用布局详解

    上回我们已经会用基本的控件创建Windows窗体应用,这才我们再来认识一些高级控件并使用ADO.NET技术连接数据库来创建功能更坚强大的窗体应用! 菜单栏控件MenuStrip .NET中提供了一个M ...