poj1163 dp入门

The Triangle

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 36811		Accepted: 22048

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. 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

5
7
3 8
8 1 0 
2 7 4 4
4 5 2 6 5

Sample Output

30

Source

IOI 1994

dp主要要找到一个初始状态，然后分解子问题，子问题最优解合起来构成要求的问题的最优解，这题是要求出从顶点走到底部，所有数字加起来和最大的一条路，我们可以把子问题看成，走过的路上的每个点到底部经过的数字的和最大，然后再分解每个点，一直分解到最后一层就OK了。程序可以从下往上写，从初始状态推最终，是我为人人的写法。

#include <iostream>
#include <cstdio>
using namespace std;

int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        int a[100][100];
        int Maxsum[100][100];
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<=i;j++)
                scanf("%d",&a[i][j]);
        }
        for(int i=0;i<n;i++)
        {
            Maxsum[n-1][i] = a[n-1][i];
        }
        for(int i=n-2;i>=0;i--)
        {
            for(int j=0;j<=i;j++)
            {
                Maxsum[i][j] =max(Maxsum[i+1][j],Maxsum[i+1][j+1])+a[i][j];
            }
        }
        printf("%d\n",Maxsum[0][0]);
    }
    return 0;
}