文章作者：姜南(Slyar) 文章来源：Slyar Home (www.slyar.com) 转载请注明，谢谢合作。
DP 数字三角形
Description
........7
......3 8
....8 1 0
..2 7 4 4
4 5 2 6 5
(Figure 1)
Figure 1 shows a number triangle. Write a program that calculates the highest sum of numbers passed on a route that starts at the top and ends somewhere on the base. Each step can go either diagonally down to the left or diagonally down to the right.
Input
Your program is to read from standard input. The first line contains one integer N: the number of rows in the triangle. The following N lines describe the data of the triangle. The number of rows in the triangle is > 1 but <= 100. The numbers in the triangle, all integers, are between 0 and 99.
Output
Your program is to write to standard output. The highest sum is written as an integer.
Sample Input
5
7
3 8
8 1 0
2 7 4 4
4 5 2 6 5
Sample Output
30
Slyar:恩，只能往下一层紧邻的2个数字走，从上到下一路过去找和最大的。转移方程如下，其中dp[i,j]表示的是从坐标[i,j]到最下面的最大和，然后只要选择下一行左边或者右边使和最大的数即可。
dp[i,j] = max(dp[i+1,j],dp[i+1,j+1])
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#include <iostream> using namespace std; const int SIZE = 101; int max(int x, int y) { return x > y ? x : y; } int main() { int n; int dp[SIZE][SIZE] = {0}; cin >> n; for (int i = 1; i <= n; i++) { for (int j = 1; j <= i; j++) { cin >> dp[i][j]; } } for (int i = n-1; i >= 1; i--) { for (int j = 1; j <= i; j++) { dp[i][j] += max(dp[i+1][j], dp[i+1][j+1]); } } cout << dp[1][1] << endl; return 0; } |
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