nyoj 18-The Triangle(动态规划)

18-The Triangle

内存限制:64MB 时间限制:1000ms Special Judge: No
accepted:5 submit:5

题目描述:

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.

输入描述:

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.

输出描述:

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

样例输入:

复制

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

样例输出:

30

分析：
　　①、因为肯定包含第一个数A[1][1]，所以为了便于理解，我们不妨从下往上思考
　　②、及就是第n-1我们就确定出对应的n-1个的最大值情况
　　③、状态方程：A[i][j] += max(A[i+1][j], A[i+1][j+1])
步骤：
　　①、从倒数第二层向上依次遍历
　　②、每一层根据状态方程算出该层每一个值对应向下可以得到的最大值
　　③、A[1][1]即为所求

核心代码：
　　

for(int i = n-1; i>=1; -- i)
    for(int j = 1; j <= i; ++ j)
        A[i][j] += max(A[i+1][j], A[i+1][j+1]);
printf("%d
", A[1][1]);

C/C++代码实现（AC）：

#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <cstdio>
#include <queue>
#include <set>
#include <map>
#include <stack>

using namespace std;
const int MAXN = 110;
int A[MAXN][MAXN];

int main ()
{
    int n;
    scanf("%d", &n);
    for(int i = 1; i <= n; ++ i)
        for (int j = 1; j <= i; ++ j)
            scanf("%d", &A[i][j]);

    for(int i = n-1; i >= 1; -- i)
    {
        for (int j = 1; j <= i; ++ j)
        {
            A[i][j] = max(A[i + 1][j], A[i + 1][j + 1]) + A[i][j];
        }
    }
    printf("%d
", A[1][1]);
    return 0;
}