杭电三部曲一、基本算法；19题 Cow Bowling

Problem 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

 

Source

PKU

题意：给定一个数塔，第i行有i个数。现在需要找到一条从树顶结点遍历到树底结点的最长路径，即遍历的结点中数字相加的和最大。且每个结点只许向其下层相邻的左右两个结点遍历。
 思路：看到题目一开始想用搜索做但是数据范围太大；如果用广搜的话，只是最后一步就需要2*10^105个空间肯定会超时，效率也不高；所以，改用dp；记录到第i行第j列是和的最大值；

 

#include<iostream>
#include<cstdio>

using namespace std;

int max(int &a,int &b)
{
    return a>b?a:b;
}

int main()
{
    int n;
    cin>>n;
    int i,j;
    int dp[353]={0};//dp数组，
    int s[353]={0};//存储数组；
    for(i=1;i<=n;i++)
    {
        for(j=1;j<=i;j++)//录入第i行
            cin>>s[j];
        for(j=i;j>0;j--)//对dp数组进行跟新；
            dp[j]=max(dp[j],dp[j-1])+s[j];
    }
    j=0;
    for(i=1;i<=n;i++)
        if(dp[i]>j)j=dp[i];//查找最大值；
    cout<<j<<endl;//输出最大值；
    return 0;
}