《程序设计与算法算法基础》《第六周 分治》动态规划
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1163:The Triangle
- 总时间限制:
- 1000ms
- 内存限制:
- 65536kB
- 描述
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.- 输入
- 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.
- 输出
- Your program is to write to standard output. The highest sum is written as an integer.
- 样例输入
-
5 7 3 8 8 1 0 2 7 4 4 4 5 2 6 5
- 样例输出
-
30
/* http://bailian.openjudge.cn/practice/1163/ 1163:The Triangle 递归解法2:数字三角形的记忆递归型动归程序 */ #include<iostream> #include<algorithm> #define MAX 101 using namespace std; int D[MAX][MAX]; int sum[MAX][MAX]; int n; int MaxSum(int i, int j) { if (sum[i][j] != -1)/*说明这个路径的最大和已经算过了*/ { return sum[i][j]; } if (i == n) { sum[i][j] = D[i][j]; } else { int x = MaxSum(i + 1, j); int y = MaxSum(i + 1, j + 1); sum[i][j] = max(x, y) + D[i][j]; } return sum[i][j]; } int main() { int i, j; cin >> n; for (i = 0; i < n; i++) { for (j = 0; j <= i; j++) { cin >> D[i][j]; sum[i][j] = -1; } } cout << MaxSum(0, 0) << endl; return 0; }
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