POJ 1163 The Triangle DP

最新推荐文章于 2022-03-18 22:07:07 发布

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最新推荐文章于 2022-03-18 22:07:07 发布

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分类专栏： Data Structure And Algorithm 文章标签： DP

本文链接：https://blog.youkuaiyun.com/qq_40624026/article/details/88426887

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该博客探讨了如何解决POJ 1163问题，即找到一个数字三角形中从顶部到底部的最大数字和。作者提出了三种解决方案：递归、记忆化搜索和循环递推，并解释了如何通过动态规划降低时间复杂度，最终实现AC（Accepted）状态。

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题目：题目链接

题目描述：
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.

题意;图1显示了一个数字三角形。编写一个程序，计算从顶部开始到底部某处的路径上传递的最大数字总和。每个步骤可以沿对角线向下滑动，也可以沿斜向下滑动到右侧。

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
5
7
3 8
8 1 0 
2 7 4 4
4 5 2 6 5
Sample Output
30

思路1：递归MaxSum（i，j） = max(MaxSum(i+1,j),MaxSum(i+1,j+1))+D[i][j];加上递归出口，if i == n MaxSum(i,j) = D[i][j](说明已经到了最后一行，直接返回即可) (TLE)

思路2;

7(1)

3 (1) 8(1)

8(1) 1(2) 0(1)

2(1) 7(3) 4(3) 4(1)

4(1) 5(4) 2(6) 6(4) 5(1)

从图中发现，第一中方法递归重复计算了很多数据，导致时间复杂度升高，为了解决重复计算的问题，采用记忆化搜索，将已经计算过的数据保存至一个数组中，当需要再次使用时，至于要直接拿出来用即可，无需重复计算，if maxSum[]i[j] != -1,说明没有计算，然后按思路1的过程计算maxSum[i][j],如果已经计算过了，直接return maxSum[i][j].(AC)

思路三：

用循环递推来计算maxSum数组，maxSum[n][i] = D[n][i](i>=1&&i<=n),maxSum[i][j] = max(maxSum[i+1][j],maxSum[i+1][j+1])+ D[i][j] (AC)

思路一代码：

/*
题目链接：https://vjudge.net/problem/POJ-1163
分类：递归 
*/
#include <iostream>
using namespace std;
const int MAX = 101;
int n;
int D[MAX][MAX];
int MaxSum(int i,int j)
{
	if(i == n) return D[i][j];//已到最后一行 
	else{
		return max(MaxSum(i+1,j),MaxSum(i+1,j+1))+D[i][j];
	} 
}
int main()
{
	cin>>n;
	for(int i = 1;i <= n;i++){
		for(int j = 1;j <= i;j++){
			cin>>D[i][j];
		}
	} 
	cout<<MaxSum(1,1)<<endl;
}

思路二代码：

/*
题目链接：https://vjudge.net/problem/POJ-1163
分类：DP 递归 
*/
#include <iostream>
using namespace std;
const int MAX = 101;
int n;
int D[MAX][MAX];
int maxSum[MAX][MAX];
int MaxSum(int i,int j)
{
	if(maxSum[i][j]!=-1) return maxSum[i][j];
	if(i == n) maxSum[i][j] = D[i][j];//已到最后一行 
	else{
		maxSum[i][j] = max(MaxSum(i+1,j),MaxSum(i+1,j+1))+D[i][j];
	} 
	return maxSum[i][j];
}
int main()
{
	cin>>n;
	for(int i = 1;i <= n;i++){
		for(int j = 1;j <= i;j++){
			cin>>D[i][j];
			maxSum[i][j] = -1;
		}
	} 
	cout<<MaxSum(1,1)<<endl;
}

思路三代码;

#include <iostream>
const int MAX = 101;
using namespace std;
int n;
int D[MAX][MAX];
int MaxSum[MAX][MAX]; 
int main(){
	cin>>n;
	for(int i = 1;i <= n;i++){
		for(int j = 1;j <= i;j++){
			cin>>D[i][j];
		}
	}
	for(int i = 1;i <= n;i++){
		MaxSum[n][i] = D[n][i];
	}
	for(int i = n-1;i >= 1;i--){
		for(int j = 1;j <= i;j++){
			MaxSum[i][j] = max(MaxSum[i+1][j],MaxSum[i+1][j+1]) + D[i][j];
		}
	}
	cout<<MaxSum[1][1]<<endl;
	return 0;
}