POJ1163 The Triangle(数字三角形) （动态规划初步）

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.InputYour 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.OutputYour 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

一个如图的数字三角形，每个点上都有些数字，从最上层开始，往下走，要么左下，要么右下，求走到最底层的路径中途径的数字和最大是多少。

这是笔者做的第一道动态规划的题，从上往下贪心肯定会出大问题。那么就从下往上。

d[i][j]表示第i层的第j个点往下走可以获得的最大和，由于只能往左下或者右下走，所以d[i][j]=max(d[i+1][j+1],d[i+1][j]),这个东西叫做状态转移方程，是动态规划的核心所在。这样相当于从下往上贪心，是不会出问题的。最后d[0][0]就是答案。这也就包含了一个“全局最优解一定包含局部最优解”的思想。

当然，dfs也是可以的，只要你电脑的CPU够强大。。。。。。。

#include <iostream>#include <stdio.h>#include <algorithm>using namespace std;int main(){int n,i,j,a[100][100]={0};scanf("%d",&n);for (i=0;i<=n-1;i++){for (j=0;j<=i;j++){scanf("%d",&a[i][j]);}}for (i=n-2;i>=0;i--){for (j=0;j<=i;j++){a[i][j]=a[i][j]+max(a[i+1][j],a[i+1][j+1]);}}printf("%d",a[0][0]);return 0;}