zoukankan      html  css  js  c++  java
  • Cow Bowling

    Cow Bowling
    Time Limit: 1000MS Memory Limit: 65536K
    Total Submissions: 15585 Accepted: 10363

    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
    简单DP;

    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <cmath>
    #include <queue>
    #include <algorithm>
    using namespace std;
    
    typedef long long LL;
    
    const int MAX =  1100;
    
    int a[400][400];
    
    int n;
    
    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]);
                }
            }
            for(int i=n-1;i>=1;i--)
            {
                for(int j=1;j<=i;j++)
                {
                    a[i][j]+=max(a[i+1][j],a[i+1][j+1]);
                }
            }
            printf("%d
    ",a[1][1]);
        }
    
        return 0;
    }
    
  • 相关阅读:
    C语言第三次博客作业---单层循环结构
    C语言第二次博客作业---分支结构
    C语言第一次博客作业——输入输出格式
    C语言--第0次作业
    Codeforces Round #341 Div.2 A. Wet Shark and Odd and Even
    Sources
    kuangbin_SegTree E (HDU 1698)
    (MST) HDOJ 1102 Constructing Roads
    kuangbin_SegTree B (HDU 1754)
    kuangbin_SegTree A (HDU 1166)
  • 原文地址:https://www.cnblogs.com/juechen/p/5255978.html
Copyright © 2011-2022 走看看