xmu 1010

Description

Consider the number triangle shown below. Write a program that calculates the highest sum of numbers that can be 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.

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

In the sample above, the route from 7 to 3 to 8 to 7 to 5 produces the highest sum: 30.

Input

The first line contains R (1 <= R <= 1000), the number of rows. Each subsequent line contains the integers for that particular row of the triangle. All the supplied integers are non-negative and no larger than 100.

Output

A single line containing the largest sum using the traversal specified.

Sample Input

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

Sample Output

30

Hint

Huge input, scanf is recommended if you are using C++.

 

如果从数字堆的顶部出发，那么每次都有两个选择（无法用贪心），要一个个的试验一次，那么时间的复杂度是O(2^k),k为数字堆的层数。换一种思路，如果从底部出发，每次都选上层相邻比较大的数跟自己相加，这种的贪心策略可以保证结果的真确性，而且复杂度为O（k^2），k为数字堆的层数。

 

#include <stdio.h>#define maxn 1005int  v[maxn][maxn];int main(){int n;int i, j;long temp;scanf("%d", &n);for (i=0; i<n; i++){for (j=0; j<i+1; j++){scanf("%ld", &v[i][j]);}}for (i=n-2; i>=0; i--){for (j=0; j<=i; j++){temp = v[i+1][j]>v[i+1][j+1] ? v[i+1][j]:v[i+1][j+1];v[i][j] += temp;}}printf("%ld\n", v[0][0]);return 0;}