uva 11044 Searching for Nessy

Searching for Nessy

The Loch Ness Monsteris a mysterious and unidentified animal said to inhabit Loch Ness,
a large deep freshwater loch near the city of Inverness in northern Scotland. Nessie is usually categorized as a type of lake monster.
http://en.wikipedia.org/wiki/Loch_Ness_Monster

In July 2003, the BBC reported an extensive investigation of Loch Ness by a BBC team, using 600 separate sonar beams, found no trace of any ¨sea monster¨ (i.e., any large animal, known or unknown) in the loch. The BBC team concluded that Nessie does not exist. Now we want to repeat the experiment.

Given a grid of n rows and m columns representing the loch, 6 uva 11044 Searching for Nessy n, m10000, find the minimum number s of sonar beams you must put in the square such that we can control every position in the grid, with the following conditions:

one sonar occupies one position in the grid; the sonar beam controls its own cell and the contiguous cells;
the border cells do not need to be controlled, because Nessy cannot hide there (she is too big).

For example,

uva 11044 Searching for Nessy

where X represents a sonar, and the shaded cells are controlled by their sonar beams; the last figure gives us a solution.

Input

The first line of the input contains an integer, t, indicating the number of test cases. For each test case, there is a line with two numbers separated by blanks, 6 uva 11044 Searching for Nessy n, m10000, that is, the size of the grid (n rows and m columns).

Output

For each test case, the output should consist of one line showing the minimum number of sonars that verifies the conditions above.

Sample Input

Sample Output

貌似题目很长很难，其实很简单，稍微思考一下就可以了，没必要过多解释。这种题目要重视，千万不要被题目的长度吓住了。这种题目是纸老虎。

拙劣的代码

 1 #include <iostream>
 2 #include <cstdlib>
 3 #include <cstdio>
 4 
 5 using namespace std;
 6 
 7 int main(void)
 8 {
 9     int n, m, t;
10 
11     cin >> t;
12     while (t--)
13     {
14         cin >> n >> m;
15         n -= 2; m-= 2;
16         n = n / 3 + (n % 3 != 0);
17         m = m / 3 + (m % 3 != 0);
18         cout << n * m << endl;
19     }
20 
21     return 0;
22 }