POJ - 3176 Cow Bowling

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

Hint

Explanation of the sample:

          7

         *

        3   8

       *

      8   1   0

       *

    2   7   4   4

       *

  4   5   2   6   5

The highest score is achievable by traversing the cows as shown above.

My Code:

#include<iostream>
#include<cstring>
#define ll long long
using namespace std;
ll way[355][355];
ll dp[355][355];
ll Max(ll x,ll y)
{
	return x>y?x:y;
}
void Input(int n)
{
	for(int i=0;i<n;i++)
	{
		for(int j=0;j<=i;j++)
		{
			cin>>way[i][j];
		}
	}
}
void Solve(int n)
{
	for(int i=0;i<n;i++)
	{
		for(int j=0;j<=i;j++)
		{
			dp[i][j]=Max(dp[i-1][j-1],dp[i-1][j])+way[i][j];
		}
	}
}
void Output(int n)
{
	int max=-1;
	for(int i=0;i<n;i++)
	{
		max=Max(dp[n-1][i],max);
	}
	cout<<max<<endl;
}

int main()
{
	memset(way,0,sizeof(way));
	memset(dp,0,sizeof(dp));
	int n;
	cin>>n;
	Input(n);
	Solve(n);
	Output(n);
	return 0;
} 

這個題的思路就是簡單dp，dp[i][j]表示的是第i行第j個保齡球的值，狀態轉移方程爲dp[i][j]=max(dp[i-1][j-1],dp[i-1][j]);
當j==0時（每行第一個）只能繼承dp[i-1][0];
遍歷最後一行即可