wanghaolin2013 @ 2024-08-22 08:09:10
#include <bits/stdc++.h>
using namespace std;
int m, n, a[202][202], s = INT_MIN;
int dx[] = {0, 1, 0, -1}, dy[] = {-1, 0, 1, 0}, c = 0;
void dfs(int x, int y, int l) {
s = max(l, s);
for (int i = 0; i < 4; i++) {
int nx = x + dx[i], ny = y + dy[i];
if (nx >= 1 && nx <= n && ny >= 1 && ny <= m && a[nx][ny] < a[x][y]) {
dfs(nx, ny, l + 1);
}
}
}
int main() {
scanf("%d %d", &n, &m);
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; j++) {
scanf("%d", &a[i][j]);
}
}
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; j++) {
dfs(i, j, 1);
}
}
printf("%d", s);
return 0;
}
by szrgjxms @ 2024-08-22 08:25:50
要不试一试记忆化搜索?!
by szrgjxms @ 2024-08-22 08:29:06
用带返回值 (int) 的 dfs 会好写很多,多开一个数组来记录就可以了。
by wanghaolin2013 @ 2024-08-22 08:41:53
能多给点提示吗?
by szrgjxms @ 2024-08-22 09:10:18
在你进去搜索的时候,你可以先判断一下,如果当前这个点(x,y)是已经“走过”的,你就返回它的值(记录的时候记录最大值)。
by szrgjxms @ 2024-08-22 09:10:53
抱歉,刚才在听课,没看见
by szrgjxms @ 2024-08-22 09:42:56
@wanghaolin2013 做出来了不?要是真的做不出来,就去学一学,点这里