POJ1163 The Triangle【DP】

The Triangle
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 58810 Accepted: 35304

Description

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.

Input

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.

Output

Your program is to write to standard output. The highest sum is written as an integer.

Sample Input

5
7
3 8
8 1 0
2 7 4 4
4 5 2 6 5

Sample Output

30

Source

IOI 1994

问题链接：POJ1163 The Triangle
问题简述：（略）
问题分析：
    动态规划问题，关键是找到状态转换方程。不过这个问题的计算过程更为容易想到。
程序说明：（略）
参考链接：（略）
题记：（略）

AC的C语言程序如下：

/* POJ1163 The Triangle */

#include <iostream>
#include <stdio.h>

using namespace std;

const int N = 100;
int n, dp[N][N];

int solve()
{
    for(int i = n - 2; i >= 0; i--)
        for(int j = 0; j <= i; j++) {
            dp[i][j] += max(dp[i + 1][j], dp[i + 1][j + 1]);
        }
    return dp[0][0];
}

int main()
{
    while(scanf("%d", &n) != EOF) {
        for(int i = 0; i < n; i++)
            for(int j = 0; j <= i; j++)
                scanf("%d", &dp[i][j]);

        printf("%d\n", solve());
    }

    return 0;
}