Michael喜欢滑雪百这并不奇怪,因为滑雪的确很刺激。可是 为了获得速度,滑的区域必须向下倾斜,而且当你滑到坡底,你不得不再次走上坡或者等待升降机来载你。Michael想知道载一个区域中最长底滑坡。区域由 一个二维数组给出。数组的每个数字代表点的高度。下面是一个例子
1 2 3 4 5
16 17 18 19 6
15 24 25 20 7
14 23 22 21 8
13 12 11 10 9
一个人可以从某个点滑向上下左右相邻四个点之一,当且仅当高度减小。在上面的例子中,一条可滑行的滑坡为24-17-16-1。当然25-24-23-…-3-2-1更长。事实上,这是最长的一条。
输入的第一行表示区域的行数R和列数C(1 <= R,C <= 100)。下面是R行,每行有C个整数,代表高度h,0<=h<=10000。
输出最长区域的长度。
5 5
1 2 3 4 5
16 17 18 19 6
15 24 25 20 7
14 23 22 21 8
13 12 11 10 9
25
package main
import "fmt"
type node struct {
line int
visit bool
}
// try to solve skiing problem of DFS
func main() {
mountain = [5][5]int{
{1, 2, 3, 4, 5},
{16, 17, 18, 19, 6},
{15, 24, 25, 20, 7},
{14, 23, 22, 21, 8},
{13, 12, 11, 10, 9},
}
max := 0
for i := 0; i < 5; i++ {
for j := 0; j < 5; j++ {
path[i][j] = dfs(i, j) + 1
if path[i][j] >= max {
max = path[i][j]
}
}
}
fmt.Printf("the longest path size is %d %v\n", max, path)
}
var fuct [5][5]node
var mountain [5][5]int
var path [5][5]int
func dfs(i, j int) int {
if fuct[i][j].visit {
return fuct[i][j].line
}
fuct[i][j].visit = true
lens := 0
tmp := 0
if i+1 <= 4 && mountain[i+1][j] < mountain[i][j] {
tmp = dfs(i+1, j) + 1
fmt.Printf("i+1 %d,j %d,tmp %d\n", i+1, j, tmp)
if tmp > lens {
lens = tmp
}
}
if i-1 >= 0 && mountain[i-1][j] < mountain[i][j] {
tmp = dfs(i-1, j) + 1
fmt.Printf("i-1 %d,j %d,tmp %d\n", i-1, j, tmp)
if tmp > lens {
lens = tmp
}
}
if j+1 <= 4 && mountain[i][j+1] < mountain[i][j] {
fmt.Printf("i %d,j+1 %d,tmp %d\n", i, j+1, tmp)
tmp = dfs(i, j+1) + 1
if tmp > lens {
lens = tmp
}
}
if j-1 >= 0 && mountain[i][j-1] < mountain[i][j] {
tmp = dfs(i, j-1) + 1
fmt.Printf("i %d,j-1 %d,tmp %d\n", i, j-1, tmp)
if tmp > lens {
lens = tmp
}
}
fuct[i][j].line = lens
return lens
}