算法问题实战策略 JUMPGAME 记忆化搜索

地址 https://algospot.com/judge/problem/read/JUMPGAME

每次我们可以选择 右或者下移动当前数字 x+num || y+num

但是遍历过于低效 会TLE

我们需要建立一个记录 记录当前的格子是否已经遍历过 如果已经遍历过直接取其记录的结果即可(当前格子是否已经遍历 遍历的结果是可以到达还是不能到达)

代码如下

 1 #include <iostream>
 2 #include <vector>
 3 
 4 using namespace std;
 5 
 6 /*
 7 2
 8 7
 9 2 5 1 6 1 4 1
10 6 1 1 2 2 9 3
11 7 2 3 2 1 3 1
12 1 1 3 1 7 1 2
13 4 1 2 3 4 1 2
14 3 3 1 2 3 4 1
15 1 5 2 9 4 7 0
16 7
17 2 5 1 6 1 4 1
18 6 1 1 2 2 9 3
19 7 2 3 2 1 3 1
20 1 1 3 1 7 1 2
21 4 1 2 3 4 1 3
22 3 3 1 2 3 4 1
23 1 5 2 9 4 7 0
24 */
25 vector<vector<int>> v(150,vector<int>(150));
26 vector<vector<int>> vis(150, vector<int>(150));
27 int n; int m;
28 
29 int jump(int x,int y)
30 {
31     if (x >= m || x<0 || y <0 || y >= m) return -1;
32     if (vis[x][y] == 1)
33         return 0;
34     vis[x][y] = 1;
35     if (x == m - 1 && y == m - 1) 
36         return 1;
37     int jumpcount = v[x][y];
38     int ret1 = jump(x + jumpcount, y);
39     int ret2 = jump(x,y+ jumpcount);
40     
41     if (ret1 == 1 || ret2 == 1)
42         return 1;
43     return -1;
44 }
45 
46 
47 void solve()
48 {
49     int x = 0, y = 0;
50 
51     int ret = jump(0,0);
52     if (ret == 1) {
53         cout << "YES" << endl;
54     }
55     else {
56         cout << "NO" << endl;
57     }
58 }
59 
60 int main()
61 {
62     cin >> n;
63     while (n--) {
64         for (int i = 0; i < 150; i++)
65             for (int j = 0; j < 150; j++)
66                 v[i][j] = 0;
67         for (int i = 0; i < 150; i++)
68             for (int j = 0; j < 150; j++)
69                 vis[i][j] = 0;
70 
71 
72         cin >> m;
73         for (int i = 0; i < m; i++) {
74             for (int j = 0; j < m; j++) {
75                 cin >> v[i][j];
76             }
77         }
78 
79         solve();
80     }
81 
82 
83     return 0;
84 }

View Code