标签：01 06 int sum tree Number numtri fout 1005

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

/*
ID: traysen1
TASK: numtri
LANG: C++
*/
#include <bits/stdc++.h>
using namespace std;

int R, tree[1005][1005], dp[1005][1005];

int DP() {
	dp[0][0] = tree[0][0];
	
	for (int i = 1; i < R; i++)
		for (int j = 0; j <= i; j++) {
			if (j == 0) {
				dp[i][j] = tree[i][j] + dp[i - 1][j];
			} else {
				dp[i][j] = tree[i][j] + max(dp[i - 1][j], dp[i - 1][j - 1]);
			}
		}
	
	int highest_sum = 0;
	for (int j = 0; j < R; j++)
		highest_sum = max(highest_sum, dp[R - 1][j]);
	
	return highest_sum;
}

int main() {
	ifstream fin("numtri.in");
	fin >> R;
	for (int i = 0; i < R; i++)
		for (int j = 0; j <= i; j++)
			fin >> tree[i][j];
	fin.close();
	
	ofstream fout("numtri.out");
	fout << DP() << endl;
	fout.close();
	return 0;
} 

标签：01,06,int,sum,tree,Number,numtri,fout,1005
来源： https://blog.csdn.net/INTEGRATOR_37/article/details/112262396

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