DP-POJ 1163 Triangle（简单数字三角形）

题目链接：

http://poj.org/problem?id=1163

The Triangle

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 45992 Accepted: 27831

Description

73   88   1   02   7   4   44   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

573 88 1 0 2 7 4 44 5 2 6 5

Sample Output

30

Source

IOI 1994

题意：数字三角形。求第一层第一个数加到最后一层两个数的最大和（要求：一个数只能加下一层和他位置相邻的两个数）

思路：最基本的DP之一，本题主要考察状态转移方程。

核心代码：

num[i][j]+=max(num[i+1][j],num[i+1][j+1]);

滚动数组：从倒数第二层开始，加最后一层；一层算完之后，回滚到上一层，知道三角形顶部（相当于从下到上地更新数字三角形，更新之后num[0][0]就是最终答案）

注释的代码，可以帮助更好的理解滚动数组的原理

 

代码：

<pre name="code" class="cpp">#include<iostream>using namespace std;int main(){   int num[200][200],n;   int i,j;   cin>>n;   for(i=0;i<n;i++){       for(j=0;j<=i;j++){           cin>>num[i][j];       }    }   for (i=n-2;i>=0;i--){       for (j=0;j<=i;j++){           num[i][j]+=max(num[i+1][j],num[i+1][j+1]);          // cout<<num[i][j]<<" ";       }       //cout<<endl;    }   cout<<num[0][0]<<endl;   return 0;}

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