题目描述

Consider the number triangle shown below. Write a program that calculates the highest sum of numbers that can be passed on a route that starts at the top and ends somewhere on the base. Each step can go either diagonally down to the left or diagonally down to the right.

          7

        3   8

      8   1   0

    2   7   4   4

  4   5   2   6   5

In the sample above, the route from 7 to 3 to 8 to 7 to 5 produces the highest sum: 30.

 

 

输入

The first line contains R (1 <= R <= 1000), the number of rows. Each subsequent line contains the integers for that particular row of the triangle. All the supplied integers are non-negative and no larger than 100.

 

输出

A single line containing the largest sum using the traversal specified.

 

样例输入

5

7

3 8

8 1 0

2 7 4 4

4 5 2 6 5

样例输出

30

分析：

我就说是个DP题，递什么归递归。

设dp[i][j]表示从底层走到达坐标[i][j]的最大值，从底向上扫，状态转移方程为dp[i][j]+=max(dp[i+1][j],dp[i+1][j+1])。

#include 
     
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             #define range(i,a,b) for(int i=a;i<=b;++i)#define rerange(i,a,b) for(int i=a;i>=b;--i)#define fill(arr,tmp) memset(arr,tmp,sizeof(arr))using namespace std;int n,dp[1005][1005];void init(){ cin>>n; range(i,1,n)range(j,1,i)cin>>dp[i][j]; rerange(i,n-1,1) range(j,1,i)dp[i][j]+=max(dp[i+1][j],dp[i+1][j+1]);}void solve(){ cout<
             
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