动态规划学习(1)-数字三角形问题

为了能掌握动态规划的求解方法，以后会做大量的DP题目，也会把必要的笔记记录下来。

题目:



用递归的方法是可以解决的，以D(i,j)表示第i行第j个数字（从1开始计算），想象用一个方格网将数字三角形装起来，下一步只能是D(i+1,j)或D(i+1,j+1)，递归的出口是D(1,1).

下面给出程序:

#include<iostream>
using namespace std;

#define MAX_NUMBER 100

int D[MAX_NUMBER][MAX_NUMBER];
int N;
int  MaxSum(int i, int j)
{
if (i == N)return D[i][j];
int Sum1 = MaxSum(i + 1, j);
int Sum2 = MaxSum(i + 1, j + 1);
if (Sum1 > Sum2)
return Sum1 + D[i][j];
else
return Sum2 + D[i][j];
}
int main()
{
cout << "Please input the rows of numbers you want to test:" << endl;
cin >> N;
for (int i = 1; i <= N; i++)
for (int j = 1; j <= i; j++)
{
cout << "row " << i << " number " << j << ":"<<endl;
cin >> D[i][j];

}
cout << endl;
cout << "The max sum of these triangle digits is :" << MaxSum(1, 1) << endl;
return 0;
}

运行结果:

Please input the rows of numbers you want to test:
5
row 1 number 1:
7
..........
row 5 number 3:
2
row 5 number 4:
6
row 5 number 5:
5

The max sum of these triangle digits is :30

这个程序非常低效，原因是递归调用时调用函数次数过多，而且每次调用时都重新计算一次之前的结果，想想递归调用时的过程(想象成一个栈),每次计算MaxSum(i,j),都得计算MaxSum(i+1,j)和MaxSum(i+1,j+1)。

解决的办法是计算了MaxSum(i,j)后就记住，下次要用时直接拿出来，不用再重复计算.可以用一个二维数组记住MaxSum的值。

修改如下:

#include<iostream>
#include<memory.h>
using namespace std;

#define MAX_NUMBER 100

int D[MAX_NUMBER][MAX_NUMBER];
int N;
int MaxSumNow[MAX_NUMBER][MAX_NUMBER];
int  MaxSum(int i, int j)
{
if (i == N)return D[i][j];
if (MaxSumNow[i + 1][j] == -1)
MaxSumNow[i + 1][j] = MaxSum(i + 1, j);
if (MaxSumNow[i + 1][j + 1] == -1)
MaxSumNow[i + 1][j + 1] = MaxSum(i + 1, j + 1);
if (MaxSumNow[i + 1][j] > MaxSumNow[i + 1][j + 1])
return MaxSumNow[i + 1][j] + D[i][j];
return MaxSumNow[i + 1][j + 1] + D[i][j];
}
int main()
{
cout << "Please input the rows of numbers you want to test:" << endl;
cin >> N;
memset(MaxSumNow, -1, sizeof(MaxSumNow));
for (int i = 1; i <= N; i++)
for (int j = 1; j <= i; j++)
{
cout << "row " << i << " number " << j << ":" << endl;
cin >> D[i][j];

}
cout << endl;
cout << "The max sum of these triangle digits is :" << MaxSum(1, 1) << endl;
return 0;
}

哈哈，已经使用了动态规划啦。将一个问题分解为子问题进行解决，为避免重复计算而保存中间结果的方法就是动态规划。能使用动态规划方法求解问题的条件是:最优解的每个局部解也是最优的。