[1_5_numtri] Space Efficiency one-dimensional array for DP_one-dimensional arraycsdn-优快云博客

本文介绍了一个算法，用于计算数三角形中从顶点到基边的最大路径和。算法通过动态规划方法逐步更新路径总和，最终得出最优路径。

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Number Triangles

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.

PROGRAM NAME: numtri

INPUT FORMAT

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.

SAMPLE INPUT (file numtri.in)

OUTPUT FORMAT

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

SAMPLE OUTPUT (file numtri.out)

We keep track (in the "best" array) of total for the best path ending in a given column of the triangle. Viewing the input, a path through the triangle always goes down or down and to the right. To process a new row, the best path total ending at a given column is the maximum of the best path total ending at that column or the one to its left, plus the number in the new row at that column. We keep only the best totals for the current row (in "best") and the previous row (in "oldbest").

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <assert.h>

#define MAXR 1000

int
max(int a, int b)
{
	return a > b ? a : b;
}

void
main(void)
{
	int best[MAXR], oldbest[MAXR];
	int i, j, r, n, m;
	FILE *fin, *fout;

	fin = fopen("numtri.in", "r");
	assert(fin != NULL);
	fout = fopen("numtri.out", "w");
	assert(fout != NULL);

	fscanf(fin, "%d", &r);

	for(i=0; i<MAXR; i++)
		best[i] = 0;

	for(i=1; i<=r; i++) {
		memmove(oldbest, best, sizeof oldbest);
		for(j=0; j<i; j++) {
			fscanf(fin, "%d", &n);
			if(j == 0)
				best[j] = oldbest[j] + n;
			else
				best[j] = max(oldbest[j], oldbest[j-1]) + n;
		}
	}

	m = 0;
	for(i=0; i<r; i++)
		if(best[i] > m)
			m = best[i];

	fprintf(fout, "%d\n", m);
	exit(0);
}