zoukankan      html  css  js  c++  java
  • POJ-3176

    Cow Bowling
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 19864   Accepted: 13172

    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.

    题意:

    求金字塔顶到底的最大路线的值。

    从n-1行开始,每次取下一行的最大,知道顶端。

    AC代码:

     1 //#include<bits/stdc++.h>
     2 #include<iostream>
     3 #include<cmath>
     4 using namespace std;
     5 
     6 int dp[400][400];
     7 
     8 int main(){
     9     ios::sync_with_stdio(false);
    10     int n;
    11     while(cin>>n&&n){
    12         for(int i=1;i<=n;i++){
    13             for(int j=1;j<=i;j++){
    14                 cin>>dp[i][j];
    15             }
    16         }
    17         for(int i=n-1;i>=1;i--){
    18             for(int j=1;j<=i;j++){
    19                 dp[i][j]=max(dp[i+1][j],dp[i+1][j+1])+dp[i][j];
    20             }
    21         }
    22         cout<<dp[1][1]<<endl;
    23     }
    24     return 0;
    25 } 
  • 相关阅读:
    wcf第3步之报文分析及原生调用
    IBatis 批量插入数据之SqlBulkCopy
    MVC前后端数据被编码
    log4Net控制台输出
    这可能是由于服务终结点绑定未使用 HTTP 协议造成的 .这还可能是由于服务器中止了 HTTP 请求上下文
    IBatis存储过程返回值
    路由学习2
    restClient访问SSL
    hibernate多对多关系配置
    hibernate 一对多操作(级联操作)
  • 原文地址:https://www.cnblogs.com/Kiven5197/p/7347748.html
Copyright © 2011-2022 走看看