POJ 3176 Cow Bowling(基础dp)

最新推荐文章于 2020-05-15 15:31:53 发布

原创最新推荐文章于 2020-05-15 15:31:53 发布 · 195 阅读

1 ·

CC 4.0 BY-SA版权

基础dp 专栏收录该内容

35 篇文章

订阅专栏

本文介绍了一种名为CowBowling的游戏算法实现，该算法通过遍历一个数值三角形来寻找从顶部到底部路径中数值之和的最大值。文章提供了一个完整的C++程序示例，演示了如何使用动态规划的方法来解决这个问题。

Cow Bowling

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions:20882		Accepted: 13811

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.

#include<cstdio>
#include<string.h>
#include<algorithm>
#include<string>
int a[400][400],dp[400];
using namespace std;
int main()
{
    int n;scanf("%d",&n);
    int i,j,k;
    memset(dp,0,sizeof(dp));
    for(i=1;i<=n;i++)
    {
        for(j=1;j<=i;j++)
            scanf("%d",&a[i][j]);
    }
    for(i=1;i<=n;i++)
    {
        for(j=i;j>=1;j--)
        {
            if(j==1)
                dp[j]=dp[j]+a[i][j];
            else if(j==i)
                dp[j]=dp[j-1]+a[i][j];
            else
            {
                dp[j]=max(dp[j]+a[i][j],dp[j-1]+a[i][j]);
            }
        }
    }
    int ma=0;
    for(i=1;i<=n;i++)
        ma=max(ma,dp[i]);
    printf("%d",ma);
    return 0;
}