poj 3176 Cow Bowling（dp基础）

最新推荐文章于 2020-07-24 18:53:48 发布

转载最新推荐文章于 2020-07-24 18:53:48 发布 · 56 阅读

本文探讨了如何在给定的三角形数列中，通过特定的路径选择，实现路径上数字之和的最大化。通过动态规划算法解决此问题，详细解释了求解过程，并提供了代码实现。

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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

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 using namespace std;
 5 #define N 356
 6 int mp[N][N];
 7 int dp[N][N];
 8 int main()
 9 {
10     int n;
11     while(scanf("%d",&n)==1){
12         memset(dp,0,sizeof(dp));
13         for(int i=1;i<=n;i++){
14             for(int j=1;j<=i;j++){
15                 scanf("%d",&mp[i][j]);
16             }
17         }
18         dp[1][1]=mp[1][1];
19         for(int i=2;i<=n;i++){
20             for(int j=1;j<=i;j++){
21                 dp[i][j]=max(dp[i-1][j-1]+mp[i][j],dp[i-1][j]+mp[i][j]);
22             }
23         }
24         int maxn=-1;
25         for(int i=1;i<=n;i++){
26             maxn=max(maxn,dp[n][i]);
27         }
28         printf("%d\n",maxn);
29     }
30     return 0;
31 }