[TopCoder] SRM 578 DIV 2, Wolf In Zoo, Solution

Problem Statement

Mr. Pasuterukun is walking along a straight road. He is cautious, because he has heard that there may be some wolves on the road.

The road consists of N sections. The sections are numbered 0 through N-1, in order. Each section of the road contains at most one wolf.

You have M additional pieces of information about the positions of the wolves. Each piece of information is an interval of the road that contains at least one wolf. More precisely, for each i between 0 and M-1, inclusive, you are given two integers left[i] and right[i] such that the sections with numbers in the range from left[i] to right[i], inclusive, contain at least one wolf in total.

You are given two String[]s L and R. The concatenation of all elements of L will be a single space separated list containing the integers left[0] through left[M-1]. R contains all the integers right[i] in the same format.

Return the number of ways in which wolves can be distributed in the sections of the road, modulo 1,000,000,007.

Definition

Class:	WolfInZooDivTwo
Method:	count
Parameters:	int, String[], String[]
Returns:	int
Method signature:	int count(int N, String[] L, String[] R)
(be sure your method is public)

Constraints

-

N will be between 1 and 300, inclusive.

-

L and R will contain between 1 and 50 elements, inclusive.

-

Each element of L and R will contain between 1 and 50 characters, inclusive.

-

Each character in L and R will be a digit ('0'-'9') or a space (' ').

-

M will be between 1 and 300, inclusive.

-

The concatenation of all elements of L will be a single space separated list of M integers. The integers will be between 0 and N-1, inclusive, and they will be given without unnecessary leading zeroes.

-

The concatenation of all elements of R will be a single space separated list of M integers. The integers will be between 0 and N-1, inclusive, and they will be given without unnecessary leading zeroes.

-

For each i, the i-th integer in L will be smaller than or equal to the i-th integer in R.

Examples

0)

5

{"0 1"}

{"2 4"}

Returns: 27

There is at least one wolf on the sections 0 through 2, and at least one wolf on the sections 1 through 4.

1)

10

{"0 4 2 7"}

{"3 9 5 9"}

Returns: 798

2)

100

{"0 2 2 7 10 1","3 16 22 30 33 38"," 42 44 49 51 57 60 62"," 65 69 72 74 77 7","8 81 84 88 91 93 96"}

{"41 5 13 22 12 13 ","33 41 80 47 40 ","4","8 96 57 66 ","80 60 71 79"," 70 77 ","99"," 83 85 93 88 89 97 97 98"}

Returns: 250671525

You must first concatenate the elements of L and only then split it into integers. The same holds for R.

3)

3

{"1"}

{"2"}

Returns: 6

The following picture shows all possible patterns.

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[Thoughts]
这道题在TopCoder上是1000分的题，所以肯定不是简单的DFS就可以解决的。尤其N的取值是在1~300之间，DFS的话，时间复杂度肯定很高。

所以，转换个思路来做就简单的多了。
以test case 2为例

10

{"0 4 2 7"}

{"3 9 5 9"}

如图所示，这个题目说到底是个集合问题。

在代码中，只要求出Cs(A)∪Cs(B)∪Cs(C)∪Cs(D)即可。Cs是取补集的意思。