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社区首页 >专栏 >【POJ】3176 - Cow Bowling(dp)

【POJ】3176 - Cow Bowling(dp)

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FishWang
发布2025-08-26 19:56:39
发布2025-08-26 19:56:39
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Cow Bowling

Time Limit: 1000MS

Memory Limit: 65536K

Total Submissions: 17279

Accepted: 11534

Description

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:

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          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

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

Sample Output

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30

Hint

Explanation of the sample:

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          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.

Source

USACO 2005 December Bronze

逆向思维从底部往顶部走,然后动态规划来选择最大路径。

代码如下:

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#include <cstdio>
#include <algorithm>
using namespace std;
int mapp[360][360];
int main()
{
	int n;
	while (~scanf ("%d",&n))
	{
		for (int i = 1 ; i <= n ; i++)
			for (int j = 1 ; j <= i ; j++)
				scanf ("%d",&mapp[i][j]);
		for (int i = n ; i >= 2 ; i--)
		{
			for (int j = 1 ; j < i ; j++)
				mapp[i-1][j] += max(mapp[i][j] , mapp[i][j+1]);
		}
		printf ("%d\n",mapp[1][1]);
	}
	return 0;
}
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原始发表:2016-07-27,如有侵权请联系 cloudcommunity@tencent.com 删除

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