Hdu2084 data Tower

Shuta

Time Limit: 1000/1000 MS (Java/others) memory limit: 32768/32768 K (Java/Others)
Total submission (s): 15753 accepted submission (s): 9399

When Problem description describes the DP algorithm, a classic example is the data tower problem, which is described as follows:

There is a number tower as shown below. It is required to go from the top layer to the bottom layer. If each step can only go to adjacent nodes, what is the maximum sum of the numbers of the nodes that pass through?

I already told you that this is a DP question. Can you AC it?

Input data first includes an integer c, indicating the number of test instances. The first line of each test instance is an integer N (1 <= n <= 100 ), the height of the tower. Next, use N rows of numbers to represent the tower. Row I has an I integer, and All integers are within the range [0, 99.

Output for each test instance, the output may be the largest sum, and the output of each instance occupies one row.

Sample Input

1573 88 1 0 2 7 4 44 5 2 6 5

 

Sample output

30

 

Source2006/1/15 ACM Program Design Final Examination

Recommendlcy

Solution: this topic is the most basic dynamic planning question. First, if the processed data is in the (I, j) vertex (where I, j is in the Number Column), the maximum value of the current vertex is treee (I, j) + max (tree (I-1, J), tree (I-1, J-1), which is applicable to any position of the number Tower except the underlying layer. During processing, the formula can be used directly from the underlying layer to the top layer, and the data in the top layer data storage memory can be output.

# Include <cstdio> # include <algorithm> using namespace STD; int main () {int tree [101] [101]; int N, T; int I, J; scanf ("% d", & T); While (t --) {scanf ("% d", & N); for (I = 0; I <N; I ++) for (j = 0; j <= I; j ++) scanf ("% d", & tree [I] [J]); for (I = n-2; I> = 0; I --) for (j = 0; j <= I; j ++) tree [I] [J] = max (tree [I + 1] [J], tree [I + 1] [J + 1]) + tree [I] [J]; // you can directly set the state equation to printf ("% d \ n", tree [0] [0]);} return 0 ;}