1.5.1 Number Triangles(dp)

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.

PROGRAM NAME: numtri

INPUT FORMAT

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.

SAMPLE INPUT (file numtri.in)

5

7

3 8

8 1 0

2 7 4 4

4 5 2 6 5

OUTPUT FORMAT

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

SAMPLE OUTPUT (file numtri.out)

30

 

代码：

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
int m[1005][1005];
int main()
{
    freopen("numtri.in","r",stdin);
    freopen("numtri.out","w",stdout);
    int r;
    cin>>r;
    for(int i=1;i<=r;i++){
        for(int j=1;j<=i;j++){
            cin>>m[i][j];
        }
    }
    for(int i=1;i<=r;i++){
        for(int j=1;j<=i;j++){
            if(j==1) m[i][j]+=m[i-1][j];
            else if(j==i) m[i][j]+=m[i-1][j-1];
            else m[i][j]+=max(m[i-1][j-1],m[i-1][j]);
        }
    }
    int maxn=0;
    for(int i=1;i<=r;i++){
        if(maxn<m[r][i]) maxn=m[r][i]; 
    }
    cout<<maxn<<endl;
    return 0;
}