POJ 1163 The Triangle

三角形路径最大和计算
本文介绍了一种解决三角形路径最大和问题的方法,通过自底向上动态规划算法计算从三角形顶部到底部的最大路径和。

The Triangle
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 40354 Accepted: 24340

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

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

Sample Output

30

自底向上的求解方法

#include <stdio.h>
int main()
{
	int n;
	scanf("%d", &n);
	int i, j;
	int tr[105][105], max[105][105];
	for(i = 1; i <= n; i++)
		for(j = 1; j <=i; j++)
			scanf("%d", &tr[i][j]);
	for(i = n, j = 1; j <= i; j++)
		max[i][j] = tr[i][j];
	for(i = n; i >= 1; i--)
		for(j = 1; j <= i; j++)
		{
			if(max[i][j] > max[i][j+1])
				max[i-1][j] = max[i][j] + tr[i-1][j];
			else
				max[i-1][j] = max[i][j+1] + tr[i-1][j];
		}
	printf("%d\n", max[1][1]);
	return 0;
}


还不太懂什么是DP,别人的代码。


#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int max[105][105], tr[105][105];
int n;
int maxsum(int x, int y)
{
	if(x == n)
		return tr[x][y];
	if(max[x+1][y] == -1)
		max[x+1][y] = maxsum(x+1, y);
	if(max[x+1][y+1] == -1)
		max[x+1][y+1] = maxsum(x+1, y+1);
	if(max[x+1][y] > max[x+1][y+1])
		return max[x+1][y] + tr[x][y];
	else
		return max[x+1][y+1] + tr[x][y];
}

int main()
{
	int i, j;
	scanf("%d", &n);
	memset(max, -1, sizeof(max));
	for(i = 1; i <= n; i++)
		for(j = 1; j <= i; j++)
			scanf("%d", &tr[i][j]);
	printf("%d", maxsum(1,1));
	return 0;
}




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