POJ 1163 The Triangle（基础DP）

解题思路：基础DP题目，使用滚动数组，只要一维数组就够了，节省空间。定义ans[]为每行前j个中的最大值，

ans[j] = max( ans[j], ans[j+1]) + t[i][j];为递推公式。 ans[1]即为t[i][j]为三角形的最大值。

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
#include<cstdio>#include<algorithm>#include<iostream>using namespace std;int t[305][305];int ans[305];int main(){    int n;    cin>>n;    for(int i = 1; i <= n ; i++)    {        for(int j = 1 ; j <= i ; j++)        {            cin>>t[i][j];        }    }    for(int i = 1; i <= n ; i++)        ans[i] = t[n][i];    for(int i = n -1 ; i >= 1 ; i--)    {        for(int j = 1 ; j <= i ; j++)        {            ans[j] = max( ans[j], ans[j+1]) + t[i][j];        }    }    cout<<ans[1]<<endl;}