NYOJ 18 The Triangle（dp + 记忆化）

最新推荐文章于 2020-09-10 11:41:06 发布

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最新推荐文章于 2020-09-10 11:41:06 发布

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本文链接：https://blog.youkuaiyun.com/x_y_q_/article/details/52087942

本文介绍了一个经典编程问题——寻找三角形数字阵列中的最大路径和，并提供了一段C++代码实现。通过动态规划方法，文章详细展示了如何从顶部开始找到到达底部某点的最大路径总和，每一步可以选择向左下或右下移动。

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The Triangle

时间限制：1000 ms | 内存限制：65535 KB

难度：4

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

样例输出

很经典的题目了这里d[t] > = 0 这个条件很关键，可以优化不少时间。它大于0代表已经搜过了，可以避免重复搜索。

<strong>#include <iostream>
#include <string.h>
#include <algorithm>
using namespace std;
int const MAX = 105;
int n, s[MAX*MAX], d[MAX*MAX];
int dp(int t, int k)
{
    if(d[t] >= 0)
        return d[t];
    return d[t] = s[t] + (k == n ? 0 : max(dp(t+k, k+1), dp(t+k+1, k+1)));
}

int main()
{
    memset(d, -1, sizeof(d));
    cin >> n;
    int m = n * (n + 1) / 2;
    for(int i = 1; i <= m; i++)
        cin >> s[i];
    cout  << dp(1,1) << '\n';
    return 0;
}
</strong>