HLG 1004 The Triangle（数塔问题DP）

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本文介绍了一个使用动态规划（DP）算法解决三角形路径最大和问题的实例。通过阅读本教程，读者可以了解如何将问题分解为更小的子问题，并通过迭代计算得出最优解。

The Triangle

点击打开题目链接

Time Limit: 1000 MS

Memory Limit: 65536 K

Total Submit: 571(236 users)

Total Accepted: 287(221 users)

Rating:

Special Judge: No

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.

For each test case, 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.

Process to the end of file.

Output

For each test case, 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

简单的DP题，求从root到达三角形底部的任意一个位置的sum和最大的值。

化为小的问题：（dp[i,j]代表达到第i行第j个数的sum的最大值）

代码：

#include <iostream>
#include<string.h>
#include<stdio.h>
#define N 110
#define max(a,b) a>b?a:b
using namespace std;
int a[N][N];
int main()
{
    int n;
    int dp[N][N],i,j,ans;
    while(~scanf("%d",&n))
    {
        for(i=0;i<n;i++)
        {
            for(j=0;j<=i;j++)
            scanf("%d",&a[i][j]);
        }
        dp[0][0]=a[0][0];
        for(i=1;i<n;i++)
        {
            dp[i][0]=dp[i-1][0]+a[i][0];
            for(j=1;j<i;j++)
            {
                dp[i][j]=max(dp[i-1][j]+a[i][j],dp[i-1][j-1]+a[i][j]);
            }
            dp[i][j]=dp[i-1][j-1]+a[i][j];
        }
        ans=0;
        for(i=0;i<n;i++)
        {
            ans=max(ans,dp[n-1][i]);
        }
        cout<<ans<<endl;
    }
    return 0;
}