POJ-3176 Cow Bowling（基础dp）

The cows don't use actual bowling balls when they go bowling. They each take a number (in the range 0..99), though, and line up in a standard bowling-pin-like triangle like this:
          7

        3   8

      8   1   0

    2   7   4   4

  4   5   2   6   5
Then the other cows traverse the triangle starting from its tip and moving "down" to one of the two diagonally adjacent cows until the "bottom" row is reached. The cow's score is the sum of the numbers of the cows visited along the way. The cow with the highest score wins that frame. 

Given a triangle with N (1 <= N <= 350) rows, determine the highest possible sum achievable.
Input

Line 1: A single integer, N 

Lines 2..N+1: Line i+1 contains i space-separated integers that represent row i of the triangle.
Output

Line 1: The largest sum achievable using the traversal rules
Sample Input

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

30

刚差不多刚看完动态规划，把小白书的例题大半都看懂了，也查了不少博客，然而做起题目来，连状态都不太会定义。

于是乎又查了起来，嗯看到了别人对状态的定义：way[i][j]表示以第i行j列的位置作为终点的路线的最大权值。 （注意区分初始化时的意义）

好的开始自己写状态转移方程：dp[i][j] = a[i][j] + max( dp[i+1][j], dp[i+1][j+1])

后来发现别人的更加简单：num[i][j] += max(num[i+1][j], num[i+1][j+1]) 

由于递推是沿着三角形向上的，就直接在原来的值上更新，只是得注意区分初始化时的意义

附上AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int num[355][355];
int main()
{
    int n;
    cin>>n;
    for (int i = 1; i <=n; i++)
        for (int j = 1; j <=i; j++)
            cin>>num[i][j];
    for (int i = n; i >= 1; i--)
        for (int j = 1; j <= i; j++)
            num[i][j] += max(num[i+1][j], num[i+1][j+1]);
    cout<<num[1][1]<<endl;

    return 0;
}