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nyoj18 The Triangle

2016-05-06 20:28 519 查看
题目来源:http://acm.nyist.net/JudgeOnline/problem.php?pid=18


The Triangle

时间限制:1000 ms  |  内存限制:65535 KB
难度:4

描述

7

3 8

8 1 0

2 7 4 4

4 5 2 6 5

(Figure 1)

Figure 1 shows a number triangle. Write a program that calculates the highest sum of numbers 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.

输入Your program is to read from standard input. The first line contains one integer N: the number of rows in the triangle. The following N lines describe the data of the triangle. The number of rows in the triangle is > 1 but
<= 100. The numbers in the triangle, all integers, are between 0 and 99.
输出Your program is to write to standard output. The highest sum is written as an integer.
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


上传者苗栋栋

//找出子问题 这题的子问题我开始以为是每一层作为一个子问题, 但是一想 影响最终结果的是全局最优解, 

//每层的最优不能作为全局最优的前提,所以子问题应该是到达最后一行每一个点的最大值 

//特别提醒下 用数组喜欢从下标0开始的同学, 为了刷题 尽量从1开始吧 很多算法中会出现 i -1这样的情况,就会访问到-1的下标

//虽然这样访问不会报错,但是我们不能让程序有隐患,而且从0开始有时候算法改起来真的麻烦,很容易考虑不全,当然只是我的感受。

 # include <stdio.h>

# include <iostream>

using namespace std;

# define MaxRow 101

int map[MaxRow][MaxRow];

int dp[MaxRow][MaxRow]; 

int main()

{
int N, i, j, sum = 0;
freopen("in.txt", "r", stdin);
scanf("%d", &N);
for(i = 1; i <= N; i++){
for(j = 1; j <= i; j++){

//因为是自底向上的递推,所以其实就是从第一层开始计算,那么直接和输入合在一起了 这样省时 排名高一点 哈哈
scanf("%d", &map[i][j]);
dp[i][j] = max(dp[i-1][j-1], dp[i-1][j]);
dp[i][j] += map[i][j];
}
}
for(i = 1; i <= N; i++){
if(sum < dp
[i]) sum = dp
[i];
}
printf("%d\n", sum);
return 0;

 } 
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