code chef - Divide the Tangerine 橘子分块算法题解

Once Chef decided to divide the tangerine into several parts. At first, he numbered tangerine's segments from1 to n in the clockwise order starting from some segment. Then he intended to divide the fruit into several parts. In order to do it he planned to separate the neighbouring segments in k places, so that he could get kparts: the 1^st - from segment l₁ to segment r₁ (inclusive), the 2^nd - from l₂ to r₂, ..., the k^th - from l_k to r_k (in all cases in the clockwise order). Suddenly, when Chef was absent, one naughty boy came and divided the tangerine into p parts (also by separating the neighbouring segments one from another): the 1^st - from segment a₁ to segment b₁, the 2^nd - from a₂ to b₂, ..., the p^th - from a_p to b_p (in all cases in the clockwise order). Chef became very angry about it! But maybe little boy haven't done anything wrong, maybe everything is OK? Please, help Chef to determine whether he is able to obtain the parts he wanted to have (in order to do it he can divide p current parts, but, of course, he can't join several parts into one).

Please, note that parts are not cyclic. That means that even if the tangerine division consists of only one part, but that part include more than one segment, there are two segments which were neighbouring in the initial tangerine but are not neighbouring in the division. See the explanation of example case 2 to ensure you understood that clarification.

Input

The first line of the input contains an integer T denoting the number of test cases. The description of T test cases follows.
The first line of each test case contains three space separated integers n, k, p, denoting the number of tangerine's segments and number of parts in each of the two divisions. The next k lines contain pairs of space-separated integers l_i and r_i. The next p lines contain pairs of space-separated integers a_i and b_i.

It is guaranteed that each tangerine's segment is contained in exactly one of the first k parts and in exactly one of the next p parts.

Output

For each test case, output a single line containing either "Yes" or "No" (without the quotes), denoting whether Chef is able to obtain the parts he wanted to have.

 

Constraints

• 1 ≤ T ≤ 100
• 1 ≤ n ≤ 5 * 10⁷
• 1 ≤ k ≤ min(500, n)
• 1 ≤ p ≤ min(500, n)
• 1 ≤ l_i, r_i, a_i, b_i ≤ n

 

Example

Input:
2
10 3 2
1 4
5 5
6 10
1 5
6 10
10 3 1
2 5
10 1
6 9
1 10

Output:
Yes
No

 

Explanation

Example case 1: To achieve his goal Chef should divide the first part (1-5) in two by separating segments 4 and 5 one from another.

Example case 2: The boy didn't left the tangerine as it was (though you may thought that way), he separated segments 1 and 10 one from another. But segments 1 and 10 are in one part in Chef's division, so he is unable to achieve his goal.

好长的题目，本题目的难度就是如何读懂题目了。

题目解析：

1 把一个橘子分块，然后分堆，所有堆中有的快的序号必须是顺时针数起

2 分堆已经乱了

3 是否在乱了的堆中分出原来想要分的堆？

额外隐藏条件： 1 所有堆都需要分出来 2 不能合并堆，只能分开堆

读懂题目就能出答案了，尤其是隐藏条件

注意：

1 输出格式Yes不是YES  

2 vector容器每次都需要清零

这个程序的时间效率有点复杂，是：O(lg(k)*max(k, p))

[cpp]  view plain copy print ?

1. #include <iostream>  
2. #include <vector>  
3. #include <string>  
4. #include <algorithm>  
5.   
6. using namespace std;  
7.   
8. struct Tangerine  
9. {  
10.     int left, right;  
11.     bool operator<(const Tangerine &t) const  
12.     {  
13.         return left < t.left;  
14.     }  
15. };  
16.   
17. bool biDiv(vector<Tangerine> &vk, int low, int up, int left)  
18. {  
19.     if (low > up) return false;  
20.     int mid = low + ((up-low)>>1);  
21.     if (vk[mid].left < left) return biDiv(vk, mid+1, up, left);  
22.     if (left < vk[mid].left) return biDiv(vk, low, mid-1, left);  
23.     return true;  
24. }  
25.   
26. string divTangerine(vector<Tangerine> &vk, vector<Tangerine> &vp)  
27. {  
28.     sort(vk.begin(), vk.end());  
29.     for (unsigned i = 0; i < vp.size(); i++)  
30.     {  
31.         if (!biDiv(vk, 0, vk.size()-1, vp[i].left)) return "No";  
32.     }  
33.     return "Yes";  
34. }  
35.   
36. void DivideTheTangerine()  
37. {  
38.     int n = 0, k = 0, p = 0;  
39.     Tangerine tang;  
40.     vector<Tangerine> vk;  
41.     vector<Tangerine> vp;  
42.   
43.     int T = 0;  
44.     cin>>T;  
45.     while (T--)  
46.     {  
47.         cin>>n>>k>>p;  
48.         for (int i = 0; i < k; i++)  
49.         {  
50.             cin>>tang.left>>tang.right;  
51.             vk.push_back(tang);  
52.         }  
53.         for (int i = 0; i < p; i++)  
54.         {  
55.             cin>>tang.left>>tang.right;  
56.             vp.push_back(tang);  
57.         }  
58.           
59.         cout<<divTangerine(vk, vp)<<endl;  
60.   
61.         vk.clear(), vp.clear();  
62.     }  
63. }