您的位置：首页 > 其它

POJ-1163 The Triangle 「数字三角形」（dfs+记忆化）

2014-04-21 20:16 295 查看

The Triangle

Time Limit: 1000MS		Memory Limit: 10000K

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

Sample Output

Source

IOI 1994

自上至下DP解法：AOJ-64 数字三角形

现在使用记忆化搜索解决它。使用递归解法的时候，因为子树的重复计算（重叠子问题），效率十分低下。对于计算过的结点进行记忆化即可。注意边界。

代码如下：

/****************************************/
#include <cstdio>
/****************************************/
const int N = 105;
int mat

, ans

;
int row;

int dfs(int i, int j) {
if(i == row)	return mat[i][j];//最后一行，返回自身。(递归边界)
if(ans[i][j])	return ans[i][j];
return ans[i][j] = max(dfs(i+1, j), dfs(i+1, j+1))+mat[i][j];//是从左边走上来的大还是右边？用大的
}

int main() {
scanf("%d", &row);
for(int i = 1; i <= row; i++) {
for(int j = 1; j <= i; j++) {
scanf("%d", &mat[i][j]);
}
}
printf("%d\n", dfs(1, 1));
return 0;
}

内容来自用户分享和网络整理，不保证内容的准确性，如有侵权内容，可联系管理员处理

标签：

相关文章推荐

新的分享

章节导航