P1101. 单词方阵

考点

  • DFS

题解

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
#include <bits/stdc++.h>
using namespace std;
const int LEN = 120;
string str = "yizhong";
int N, walk[8][2] = {{0, 1}, {0, -1}, {-1, 0}, {1, 0}, {1, 1}, {-1, 1}, {1, -1}, {-1, -1}};
char arr[LEN][LEN], ans[LEN][LEN];

bool valid(int x, int y)
{
return (x >= 1 && x <= N && y >= 1 && y <= N);
}

//虚假的dfs,直接逐点枚举八个方向即可
void dfs(int ux, int uy)
{
for (int i = 0; i < 8; ++i)
{
int diff_x = walk[i][0], diff_y = walk[i][1];
int x = ux, y = uy, j = 1;
while (j < 7 && valid(x += diff_x, y += diff_y) && str[j] == arr[x][y])
++j;
if (j < 7)
continue;
for (j = 0, x = ux, y = uy; j < 7; ++j, x += diff_x, y += diff_y)
ans[x][y] = arr[x][y];
}
}

int main()
{
cin >> N;
for (int i = 1; i <= N; ++i)
{
for (int j = 1; j <= N; ++j)
{
cin >> arr[i][j];
ans[i][j] = '*';
}
}
for (int i = 1; i <= N; ++i)
{
for (int j = 1; j <= N; ++j)
{
if (arr[i][j] == 'y')
dfs(i, j);
}
}
for (int i = 1; i <= N; ++i, cout << endl)
{
for (int j = 1; j <= N; ++j)
{
cout << ans[i][j];
}
}
return 0;
}

思路

虚假的DFS.....每次找到y字母,就向8个方向遍历即可.....

如果找到了yizhong字符串,就复制到对应位置的答案数组保存