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NYOJ - The Triangle(经典dp)

热度:42   发布时间:2024-01-20 18:03:27.0

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
5
7
3 8
8 1 0 
2 7 4 4
4 5 2 6 5
样例输出
30

这个是经典的dp问题,可以从后面向前递推,那么,动态转移方程为a[i][j] += max(a[i+1][j],a[i+1],[j+1]),下面是具体的代码,这里提供参考。

#include <stdio.h>int max(int n,int m)
{return n > m ? n : m;
}int main()
{int a[101][101];int n,i,j;while(scanf("%d",&n) != EOF){for(i = 0; i < n; i++){for(j = 0; j <= i; j++){scanf("%d",&a[i][j]);}}for(i = n-2; i >= 0; i--){for(j = 0; j <= i; j++){a[i][j] += max(a[i+1][j],a[i+1][j+1]);}}printf("%d\n",a[0][0]);}return 0;
}
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