The Triangle
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 36811 | Accepted: 22048 |
Description
7
3 8
8 1 0
2 7 4 4
4 5 2 6 5
(Figure 1)
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
dp主要要找到一个初始状态,然后分解子问题,子问题最优解合起来构成要求的问题的最优解,这题是要求出从顶点走到底部,所有数字加起来和最大的一条路,我们可以把子问题看成,走过的路上的每个点到底部经过的数字的和最大,然后再分解每个点,一直分解到最后一层就OK了。程序可以从下往上写,从初始状态推最终,是我为人人的写法。
1 #include <iostream> 2 #include <cstdio> 3 using namespace std; 4 5 int main() 6 { 7 int n; 8 while(scanf("%d",&n)!=EOF) 9 { 10 int a[100][100]; 11 int Maxsum[100][100]; 12 for(int i=0;i<n;i++) 13 { 14 for(int j=0;j<=i;j++) 15 scanf("%d",&a[i][j]); 16 } 17 for(int i=0;i<n;i++) 18 { 19 Maxsum[n-1][i] = a[n-1][i]; 20 } 21 for(int i=n-2;i>=0;i--) 22 { 23 for(int j=0;j<=i;j++) 24 { 25 Maxsum[i][j] =max(Maxsum[i+1][j],Maxsum[i+1][j+1])+a[i][j]; 26 } 27 } 28 printf("%d ",Maxsum[0][0]); 29 } 30 return 0; 31 }