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:
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:
1 for(int i = n-1; i>=1; -- i) 2 for(int j = 1; j <= i; ++ j) 3 A[i][j] += max(A[i+1][j], A[i+1][j+1]); 4 printf("%d\n", A[1][1]);
C/C++代码实现(AC):
1 #include <iostream> 2 #include <algorithm> 3 #include <cmath> 4 #include <cstring> 5 #include <cstdio> 6 #include <queue> 7 #include <set> 8 #include <map> 9 #include <stack> 10 11 using namespace std; 12 const int MAXN = 110; 13 int A[MAXN][MAXN]; 14 15 int main () 16 { 17 int n; 18 scanf("%d", &n); 19 for(int i = 1; i <= n; ++ i) 20 for (int j = 1; j <= i; ++ j) 21 scanf("%d", &A[i][j]); 22 23 for(int i = n-1; i >= 1; -- i) 24 { 25 for (int j = 1; j <= i; ++ j) 26 { 27 A[i][j] = max(A[i + 1][j], A[i + 1][j + 1]) + A[i][j]; 28 } 29 } 30 printf("%d\n", A[1][1]); 31 return 0; 32 }