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[51], R[51];

int main()

{

      ios_base::sync_with_stdio( false );

      cin.tie( 0 );

      L[1] = 1, R[1] = 1;

      for( int i = 2; i <= 50; ++i ) L[i] = L[i - 1] << 1, R[i] = ( L[i] << 1 ) - 1;

      int h, q;

      cin >> h >> q;

      if( q == 0 )

      {

            if( h == 1 ) puts( "1" );

            else puts( "Data not sufficient!" );

            return 0;

      }

      map<LL, int> mp;

      for( int i = 0; i < q; ++i )

      {

            LL level, left, right, type, gap;

            cin >> level >> left >> right >> type;

            gap = h - level;

            while( gap )

            {

                  gap--;

                  left <<= 1;

                  right = ( ( right + 1 ) << 1 ) - 1;

            }

            if( type )

            {

                  mp[left]++;

                  mp[right + 1]--;

            }

            else

            {

                  mp[L[h]]++;

                  mp[left]--;

                  mp[right + 1]++;

                  mp[R[h] + 1]--;

            }

      }

      LL ans, sum = 0, ans_gap = 0, mid_pre = -1;

      map<LL, int>:: iterator it = mp.begin();

      for( ; it != mp.end(); ++it )

      {

            sum += ( it->second );

            if( mid_pre != -1 )

            {

                  ans_gap += ( it->first ) - mid_pre;

                  ans = mid_pre;

            }

            if( sum == q ) mid_pre = ( it->first );

            else mid_pre = -1;

      }

      if( ans_gap == 1 ) cout << ans << endl;

      else if( ans_gap > 1 ) puts( "Data not sufficient!\n" );

      else puts( "Game cheated!\n" );

      return 0;

}

/*



*/