by starting at the top of the triangle below and moving to adjacent numbers on the row below, the maximum total from top to bottom is 23.
3
7 4
2 4 6
8 5 9 3
that is, 3 + 7 + 4 + 9 = 23.
find the maximum total from top to bottom of the triangle below:
75
95 64
17 47 82
18 35 87 10
20 04 82 47 65
19 01 23 75 03 34
88 02 77 73 07 63 67
99 65 04 28 06 16 70 92
41 41 26 56 83 40 80 70 33
41 48 72 33 47 32 37 16 94 29
53 71 44 65 25 43 91 52 97 51 14
70 11 33 28 77 73 17 78 39 68 17 57
91 71 52 38 17 14 91 43 58 50 27 29 48
63 66 04 68 89 53 67 30 73 16 69 87 40 31
04 62 98 27 23 09 70 98 73 93 38 53 60 04 23
最簡單的方法就是窮舉,從根節點出發,每個節點都有兩個分叉,到達底部的路徑有估計有2的指數級的數目(有會算的朋友請留言,我的組合數學都還給老師了),不過這道題顯然是符合動态規劃的特征,往下遞增一層的某個節點的最佳結果f[i][j]肯定是上一層兩個入口節點對應的最佳結果的最大值,也就是f[i-1][j]或者f[i-1][j+1],遞歸的邊界就是定點f[0][0]=75。是以我的解答如下,考慮了金字塔邊界的情況,資料按照金字塔型存儲在numbers.txt中,
import java.io.bufferedreader;
import java.io.inputstreamreader;
public class euler18problem {
public static void maxsun(int[][] a, int rows, int cols) {
// 結果清單
int[][] f = new int[15][15];
// 路徑,用于輸出計算路徑
int[][] path = new int[15][15];
// 遞歸邊界
f[0][0] = a[0][0];
path[0][0] = 0;
// 遞推
for (int i = 1; i < rows; i++) {
int col = i + 1;
// 決策
for (int j = 0; j < col; j++) {
// 左邊界
if (j - 1 < 0) {
f[i][j] = f[i - 1][j] + a[i][j];
path[i][j] = j;
} else if (j + 1 > col) { // 右邊界
f[i][j] = f[i - 1][j - 1] + a[i][j];
path[i][j] = j - 1;
} else {
// 處于中間位置
if (f[i - 1][j] <= f[i - 1][j - 1]) {
f[i][j] = f[i - 1][j - 1] + a[i][j];
path[i][j] = j - 1;
} else {
f[i][j] = f[i - 1][j] + a[i][j];
path[i][j] = j;
}
}
}
}
// 求出結果
int result = 0, col = 0;
for (int i = 0; i < cols; i++) {
if (f[14][i] > result) {
result = f[14][i];
col = i;
// 輸出路徑
system.out.println("row=14,col=" + col + ",value=" + a[14][col]);
for (int i = rows - 2; i >= 0; i--) {
col = path[i][col];
system.out.println("row=" + i + ",col=" + col + ",value="
+ a[i][col]);
system.out.println(result);
}
public static void main(string[] args) throws exception {
int rows = 15;
int cols = 15;
int[][] a = new int[rows][cols];
bufferedreader reader = new bufferedreader(new inputstreamreader(
euler18problem.class.getresourceasstream("/numbers.txt")));
string line = null;
int row = 0;
while ((line = reader.readline()) != null && !line.trim().equals("")) {
string[] numbers = line.split(" ");
for (int i = 0; i < numbers.length; i++) {
a[row][i] = integer.parseint(numbers[i]);
row++;
reader.close();
maxsun(a, rows, cols);
}
執行結果如下,包括了路徑輸出:
row=14,col=9,value=93
row=13,col=8,value=73
row=12,col=7,value=43
row=11,col=6,value=17
row=10,col=5,value=43
row=9,col=4,value=47
row=8,col=3,value=56
row=7,col=3,value=28
row=6,col=3,value=73
row=5,col=2,value=23
row=4,col=2,value=82
row=3,col=2,value=87
row=2,col=1,value=47
row=1,col=0,value=95
row=0,col=0,value=75
1074
ps.并非我閑的蛋疼在半夜做題,隻是被我兒子折騰的無法睡覺了,崩潰。
文章轉自莊周夢蝶 ,原文釋出時間2009-09-27