BYTESM2 - Philosophers Stone
One of the secret chambers in Hogwarts is full of philosopher’s stones. The floor of the chamber is covered by h × w square tiles, where there are h rows of tiles from front (first row) to back (last row) and w columns of tiles from left to right. Each tile has 1 to 100 stones on it. Harry has to grab as many philosopher’s stones as possible, subject to the following restrictions:
- He starts by choosing any tile in the first row, and collects the philosopher’s stones on that tile. Then, he moves to a tile in the next row, collects the philosopher’s stones on the tile, and so on until he reaches the last row.
- When he moves from one tile to a tile in the next row, he can only move to the tile just below it or diagonally to the left or right.
The first line consists of a single integer T, the number of test cases. In each of the test cases, the first line has two integers. The first integer h (1 <= h <= 100) is the number of rows of tiles on the floor. The second integer w (1 <= w <= 100) is the number of columns of tiles on the floor. Next, there are h lines of inputs. The i-th line of these, specifies the number of philosopher’s stones in each tile of the i-th row from the front. Each line has w integers, where each integer m (0 <= m <= 100) is the number of philosopher’s stones on that tile. The integers are separated by a space character.
The output should consist of T lines, (1 <= T <= 100), one for each test case. Each line consists of a single integer, which is the maximum possible number of philosopher’s stones Harry can grab, in one single trip from the first row to the last row for the corresponding test case.
Input: 1 6 5 3 1 7 4 2 2 1 3 1 1 1 2 2 1 8 2 2 1 5 3 2 1 4 4 4 5 2 7 5 1 Output: 32 //7+1+8+5+4+7=32
pay attention in the input, the problem says there are W integers per line, which is not true. For C/C++, reading the input is easy, since it gets the "next available int". For Python, it is not, you should read H * W integer values and then fill the "floor" matrix.
AC in one go !!
AC!! DP O(n^2)
AC in 1 go!!!!
Solved without any knowledge of dp. No recursions. Think simple.
easy problem....my 3rd dp and 1st without help
My 25th and first dp problem! :-D
MY FIRST DP ! AC IN ONE GO :)
BOTTOM UP approach
|Added by:||Paritosh Aggarwal|
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