poj 3176 Cow Bowling 【水题】

原创于 2018-05-24 23:09:40 发布 · 330 阅读

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本文介绍了一种使用动态规划解决寻找三角形网格中从顶点到底边的最大路径和问题的方法。通过逐步构建DP表来确定最高得分路径，并提供了一个C++实现示例。

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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:

The highest score is achievable by traversing the cows as shown above.

很明显，考察动态规划得知识，找出状态转移方程 dp[i][j]=max(dp[i+1][j],dp[i+1][j+1])+a[i][j]；

#include<iostream>
#include<string.h>
#include<algorithm>
using namespace std;
int a[400][400];
int dp[400][400];
int main (){
    ios::sync_with_stdio(false);
	int n,i,j;
	cin>>n;
	memset(dp,0,sizeof(dp));
	for(i=1;i<=n;i++){
		for(j=1;j<=i;j++)
		  cin>>a[i][j];
	}
	for(int i=1;i<=n;i++)
        dp[n][i] = a[n][i];  //将最后一行的dp值先用对应的a数组表示
	for(i=n-1;i>=1;i--){
		for(j=1;j<=i;j++){
			dp[i][j]=max(dp[i+1][j],dp[i+1][j+1])+a[i][j]; //逐层往上求值，最后的dp[1][1]就是答案
		}
	}
	cout<<dp[1][1]<<endl;
	return 0;
}