poj 1163

这道题是一道简单的dp题，就是求从第一个数到底行中的某个数的路径数字之和最大值。这道题的子问题就是求到每一行的每一个数的路径的最大数字和。

设：t[i][j]储存第i行第j列的数字。

maxsum[i][j]储存到第i行第j列数字的路径的最大数字和。

dp方程是：

maxsum[i][j] = (maxsum[i - 1][j - 1] + t[i][j]) > (maxsum[i - 1][j] + t[i][j]) ? (maxsum[i - 1][j - 1] + t[i][j])
: (maxsum[i - 1][j] + t[i][j]);

ps：这道题在poj上提交AC，但是在百练是提交内存溢出。

import java.util.Scanner; public class Main1163 { public static void main(String[] args) { Scanner sc = new Scanner(System.in); int n = sc.nextInt(); int[][] t = new int[n + 1][]; int[][] maxsum = new int[n + 1][]; for (int i = 1; i <= n; ++i) { t[i] = new int[i+1]; maxsum[i] = new int[i+1]; for (int j = 1; j <= i; ++j) { t[i][j] = sc.nextInt(); } } for (int i = 1; i <= n; ++i) { for (int j = 1; j <= i; ++j) { if (i - 1 == 0) maxsum[1][1] = t[1][1]; else if (j - 1 == 0) maxsum[i][j] = maxsum[i - 1][j] + t[i][j]; else if (j > i - 1) maxsum[i][j] = maxsum[i - 1][j - 1] + t[i][j]; else { maxsum[i][j] = (maxsum[i - 1][j - 1] + t[i][j]) > (maxsum[i - 1][j] + t[i][j]) ? (maxsum[i - 1][j - 1] + t[i][j]) : (maxsum[i - 1][j] + t[i][j]); } } } int max = -1; for(int i = 1 ; i <= n ; ++i){ if(maxsum[n][i] > max) max = maxsum[n][i]; } System.out.println(max); } }