【poj 3176】 Cow Bowling 递推dp

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本文介绍了一道名为 CowBowling 的编程题，题目要求从一个数字三角形的顶部开始，按照规则向下遍历到达底部，求遍历路径上的数字之和的最大值。文章给出了使用递推算法解决该问题的 C++ 代码实现。

Cow Bowling
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 16562 Accepted: 11036

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

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

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

题目链接：http://poj.org/problem?id=3176
题目大意：数字三角形；
思路：递推
代码：

    #include<iostream>
    #include<stdio.h>
    #include<string.h>
    #include<algorithm>
    using namespace std;
    int a[351][351];
    int n;
    int dp[351][351];
    int main()
    {
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
        for(int j=1;j<=i;j++)
        scanf("%d",&a[i][j]);

        dp[1][1]=a[1][1];
        for(int i=2;i<=n;i++)
        for(int j=1;j<=i;j++)
        {
            if(j!=i)
            dp[i][j]=max(dp[i-1][j],dp[i-1][j-1]);
            else dp[i][j]=dp[i-1][j-1];
            dp[i][j]+=a[i][j];
        }
        int ans=0;
        for(int i=1;i<=n;i++)
        ans=max(ans,dp[n][i]);
        printf("%d\n",ans);

    }