POJ 1163 The Triangle

最新推荐文章于 2019-01-09 16:39:02 发布

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The Triangle

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

Sample Output

题目大意：给出一个数字三角形，自顶向下找出一条路径，使得路径上的数字相加之后得到的和最大。

解题思路：经典的动态规划问题，定义dp[i][j]为自底向上的路径中到a[i][j]的最大值，则dp[i][j] = max(dp[i + 1][j + 1],dp[i + 1][j]) + a[i][j]，最后打印dp[0][0]的值即可。

代码如下：

#include <cstdio>
#include <algorithm>
#include <cstring>

using namespace std;

const int maxn = 105;
int dp[maxn][maxn],a[maxn][maxn];

int main()
{
	int t,n,i,j,k,l;
	while(scanf("%d",&n) != EOF){
		memset(dp,0,sizeof(dp));
		for(j = 0;j < n;j++)
			for(k = 0;k <= j;k++)
				scanf("%d",&a[j][k]);
		for(j = 0;j < n;j++)
			dp[n][j] = a[n][j];
		for(j = n - 1;j >= 0;j--)
			for(k = 0;k <= j;k++)
				dp[j][k] = max(dp[j + 1][k],dp[j + 1][k + 1]) + a[j][k];
		printf("%d\n",dp[0][0]);
	}
	return 0;
 }