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The Triangle

运维开发网 https://www.qedev.com 2020-06-13 16:59 出处:网络
Description 738810274445265(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

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

5
7
3 8
8 1 0 
2 7 4 4
4 5 2 6 5

Sample Output30方法:动态规划:Step 1:确定状态:

题目求解(1,1)到最底层路径的最大权值,路径起点固定,终点和中间点不确定,因此定义dg[x][y]表示从(1,1)出发到(x,y)路径的最大权值和。

最终答案就是寻找最底层的最大值,ans=max{dp[n][1],dp[n][2],dp[n][3]...dp[n][n]}.

Step 2:确定状态转移方程和边界条件:

不去考虑(1,1)到(x,y)中间是怎么走的,只需要考虑(x,y)上一步是怎么来的,上一步可能是(x-1,y),也可能是(x-1,y-1),因此dg[x][y]的最大值就是上一步的最大权值和:max{dp[x-1][y],dp[x-1][y-1]),再加上自己的权值a[x][y]。

所以状态转移方程就是:dp[x][y]=max{dp[x-1][y],dp[x-1][y-1]}+a[i][j]。

与递归一样,我们也需要终止条件,防止无限递归下去。我们发现dp[x][y]的值取决于dp[x-1][y],dp[x-1][y-1],随着递归的深入,最后一定会递归到dp[1][1],dp[1][1]不能再使用状态转移方程了,所以给dp[1][1]附一个初值,即a[1][1]。

my codes:

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 int a[550][550],dp[550][550];
 4 int n;
 5 int main()
 6 {
 7     cin>>n;
 8     for(int i=1;i<=n;i++)
 9         for(int j=1;j<=n;j++)
10             cin>>a[i][j];
11     dp[1][1]=a[1][1];
12     for(int i=2;i<=n;i++)
13     {
14         for(int j=1;j<=i;j++)
15         {
16             dp[i][j]=max(dp[i-1][j-1],dp[i-1][j])+a[i][j];
17         }
18     }
19     int ans=0;
20     for(int i=1;i<=n;i++) ans=max(ans,dp[n][i]);
21     cout<<ans<<endl;
22     return 0;
23 }
 

 更进一步的探索:https://blog.csdn.net/weixin_40851250/article/details/83622717

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