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POJ_3176_Cow_Bowling_(数字三角形)_(动态规划)

分类: IT文章 • 2023-04-02 18:56:32

描述

http://poj.org/problem?id=3176

给出一个三角形,每个点可以走到它下面两个点,将所有经过的点的值加起来,问最大的和是多少.

Cow Bowling
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 16826   Accepted: 11220

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

30

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

分析

记忆化搜索:

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 using std :: max;
 5 
 6 const int maxn=355;
 7 int a[maxn][maxn],f[maxn][maxn];
 8 int n;
 9 
10 int dfs(int i,int j)
11 {
12     if(f[i][j]!=-1) return f[i][j];
13     if(i==n) return f[i][j]=a[i][j];
14     return f[i][j]=a[i][j]+max(dfs(i+1,j),dfs(i+1,j+1));
15 }
16 
17 void init()
18 {
19     scanf("%d",&n);
20     for(int i=1;i<=n;i++)
21     {
22         for(int j=1;j<=i;j++)
23         {
24             scanf("%d",&a[i][j]);
25         }
26     }
27     memset(f,-1,sizeof(f));
28 }
29 
30 int main()
31 {
32 #ifndef ONLINE_JUDGE
33     freopen("cow.in","r",stdin);
34     freopen("cow.out","w",stdout);
35 #endif
36     init();
37     printf("%d
",dfs(1,1));
38 #ifndef ONLINE_JUDGE
39     fclose(stdin);
40     fclose(stdout);
41 #endif
42     return 0;
43 }
View Code

动态规划:

 1 #include<cstdio>
 2 #include<algorithm>
 3 using std :: max;
 4 
 5 const int maxn=355;
 6 int n;
 7 int a[maxn][maxn],f[maxn][maxn];
 8 
 9 void solve()
10 {
11     for(int i=n-1;i>=1;i--)
12     {
13         for(int j=1;j<=i;j++)
14         {
15             f[i][j]=max(f[i+1][j],f[i+1][j+1])+a[i][j];
16         }
17     }
18     printf("%d
",f[1][1]);
19 }
20 
21 void init()
22 {
23     scanf("%d",&n);
24     for(int i=1;i<=n;i++)
25     {
26         for(int j=1;j<=i;j++)
27         {
28             scanf("%d",&a[i][j]);
29         }
30     }
31     for(int j=1;j<=n;j++) f[n][j]=a[n][j];
32 }
33 
34 int main()
35 {
36 #ifndef ONLINE_JUDGE
37     freopen("cow.in","r",stdin);
38     freopen("cow.out","w",stdout);
39 #endif
40     init();
41     solve();
42 #ifndef ONLINE_JUDGE
43     fclose(stdin);
44     fclose(stdout);
45 #endif
46     return 0;
47 }
View Code

 

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