POJ3176-Cow Bowling

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探讨 CowBowling 游戏中的策略问题，通过遍历三角形结构寻找从顶到底的最大得分路径。采用动态规划算法解决，适用于算法竞赛及编程挑战。

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Cow Bowling

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 18938		Accepted: 12594

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:

          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

Sample Output

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.

Source

USACO 2005 December Bronze

题意：输入一个n层的三角形，第i层有i个数，求从第1层到第n层的所有路线中，权值之和最大为多少，第i层的某个数只能连线走到第i+1层中与它位置相邻的两个数中的一个

解题思路：dp

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <queue>
#include <cmath>
#include <map>
#include <bitset>
#include <set>
#include <vector>
#include <functional>

using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;

int a[400][400],n,dp[400][400];

int main()
{
	while (~scanf("%d", &n))
	{
		for (int i = 1; i <= n; i++)
			for (int j = 1; j <= i; j++)
				scanf("%d", &a[i][j]);
		memset(dp, 0, sizeof dp);
		for (int i = 1; i <= n; i++) dp[n][i] = a[n][i];
		for (int i = n - 1; i >= 1; i--)
			for (int j = 1; j <= i; j++)
				dp[i][j] = max(dp[i + 1][j], dp[i + 1][j + 1]) + a[i][j];
		printf("%d\n", dp[1][1]);
	}
	return 0;
}