POJ-4003 The Triangle (线性dp，入门题)

最新推荐文章于 2024-08-29 12:01:12 发布

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最新推荐文章于 2024-08-29 12:01:12 发布

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分类专栏：动态规划-线性dp 文章标签： POJ 线性dp 动态规划 ACM

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本文介绍了一个经典的算法问题——三角形路径最大和问题。该问题是寻找从三角形顶点到底边的最大数值路径。文章提供了完整的C++代码实现，并解释了动态规划方法的应用。

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The Triangle

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 51932		Accepted: 31380

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

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
int a[102][102], dp[102];
int main(){
	int n;
	scanf("%d", &n);
	for(int i = 1; i <= n; ++i){
		for(int j = 1; j <= i; ++j){
			scanf("%d", &a[i][j]);
		}
	}
	memset(dp, 0, sizeof(dp));
	for(int i = n; i >= 0; --i){
		for(int j = 1; j <= n; ++j){
			dp[j] = max(dp[j], dp[j + 1]) + a[i][j];
		}
	}
	printf("%d\n", dp[1]);	
}

/*
题意:
一个三角形，求从定点到底部的最大和路径。

思路：
线性dp，dp[j]表示某行第j个位置向下的最大和。滚动一下减少空间开销。
*/