POJ3176 Cow Bowling

Ethan_H

于 2018-05-27 17:05:55 发布

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分类专栏：动态规划 DP 文章标签：动态规划 DP 算法

本文链接：https://blog.youkuaiyun.com/qq_37755550/article/details/80470684

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在一场别开生面的保龄球比赛中，奶牛们通过选择路径累计得分来决出胜者。每只奶牛从三角形顶部出发，向下走到两个相邻的下一层奶牛中的一个，直到达到底部。任务是找到能够获得最高总分的路径。

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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.

解题思路

动态规划的入门级题目，一道数塔题目，题目分析时可以自顶向下，但解题时自底向上的效果会更好，这题的转移方程为a[i][j] += max(a[i + 1][j],a[i + 1][j + 1]);

#include<algorithm>
#include<iostream>
using namespace std;
int a[10005][10005];
int main(){
	int n;
	while(cin>>n){
		for(int i = 1;i <= n;i++){
			for(int j = 1;j <= i;j++){
				cin>>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]);
			}
		}
		cout<<a[1][1]<<endl;
	}
	return 0;
}