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

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Mean:

一棵满二叉树，树中某个叶子节点是出口，目的是寻找这个出口。再给定Q个询问的结果，每个结果告诉我们在第i层中(l,r)覆盖的叶结点是否包含出口。

analyse:

基本思路：多个区间求交集。

具体实现：

对于每一个询问，把它转化到最底层。并且把不在(l,r)区间的询问改为在(最左边，l-1)和(r+1,最右边)的形式，这样一来全部都变成了在(l,r)区间的描述。

区间统计：

对左右区间起点和终点组成的集合进行排序。然后找到答案存在的区间，如果区间长度=1，答案唯一；长度>1，答案不唯一;长度=0，无解。

Trick:会爆int。

Time complexity: O(n)

Source code: 

/*
* this code is made by crazyacking
* Verdict: Accepted
* Submission Date: 2015-07-16-11.55
* Time: 0MS
* Memory: 137KB
*/
#include <queue>
#include <cstdio>
#include <set>
#include <string>
#include <stack>
#include <cmath>
#include <climits>
#include <map>
#include <cstdlib>
#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
#define  LL long long
#define  ULL unsigned long long
using namespace std;
LL L[], R[];
int main()
{
      ios_base::sync_with_stdio( false );
      cin.tie(  );
      L[] = , R[] = ;
      for( int i = ; i <= ; ++i ) L[i] = L[i - ] << , R[i] = ( L[i] <<  ) - ;
      int h, q;
      cin >> h >> q;
      if( q ==  )
      {
            if( h ==  ) puts( "" );
            else puts( "Data not sufficient!" );
            return ;
      }
      map<LL, int> mp;
      for( int i = ; i < q; ++i )
      {
            LL level, left, right, type, gap;
            cin >> level >> left >> right >> type;
            gap = h - level;
            while( gap )
            {
                  gap--;
                  left <<= ;
                  right = ( ( right +  ) <<  ) - ;
            }
            if( type )
            {
                  mp[left]++;
                  mp[right + ]--;
            }
            else
            {
                  mp[L[h]]++;
                  mp[left]--;
                  mp[right + ]++;
                  mp[R[h] + ]--;
            }
      }
      LL ans, sum = , ans_gap = , mid_pre = -;
      map<LL, int>:: iterator it = mp.begin();
      for( ; it != mp.end(); ++it )
      {
            sum += ( it->second );
            if( mid_pre != - )
            {
                  ans_gap += ( it->first ) - mid_pre;
                  ans = mid_pre;
            }
            if( sum == q ) mid_pre = ( it->first );
            else mid_pre = -;
      }
      if( ans_gap ==  ) cout << ans << endl;
      else if( ans_gap >  ) puts( "Data not sufficient!\n" );
      else puts( "Game cheated!\n" );
      return ;
}
/*

*/