P1101 单词方阵

1. 题目链接P1101 单词方阵 - 洛谷 | 计算机科学教育新生态 (luogu.com.cn)

#include <bits/stdc++.h>
using namespace std;
#define endl '\n'
#define int long long int
int xx[] = {1,1,1,0,0,-1,-1,-1};
int yy[] = {1,0,-1,1,-1,1,0,-1};
int vis[110][110];
char a[110][110];
int n;
string yz = { "yizhong" };
void dfs(int x,int y) {//因为xx从0开始，所以i也一样//遍历8个方向，7个字符for (int i = 0; i < 8; i++) {int flag = 1;for (int j = 0; i < 7; j++) {int dx = x + j * xx[i];int dy = y + j * yy[i];if (dx<1 || dx>n || dy<1 || dy>n || yz[j] != a[dx][dy]) {flag = 0;break;}if (flag) {vis[dx][dy] = 1;}}}
}
signed main() {cin >> n;for (int i = 1; i <= n; i++) {for (int j = 1; j <= n; j++) {cin >> a[i][j];}}for (int i = 1; i <= n; i++) {for (int j = 1; j <= n; j++) {if (a[i][j] == 'y')dfs(i, j);}}for (int i = 1; i <= n; i++) {for (int j = 1; j <= n; j++) {if (!vis[i][j])cout << '*';elsecout << a[i][j];}cout << endl;}return 0;
}

2.

#include <bits/stdc++.h>
using namespace std;
#define endl '\n'
#define int long long int
int xx[] = { 1,1,1,0,0,-1,-1,-1 };
int yy[] = { 1,0,-1,1,-1,1,0,-1 };
int vis[110][110];
char a[110][110];
int n;
string yz = { "yizhong" };
void dfs(int x, int y) {//因为xx从0开始，所以i也一样//遍历8个方向，7个字符for (int i = 0; i < 8; i++) {int flag = 1;for (int j = 0; i < 7; j++) {int dx = x + j * xx[i];int dy = y + j * yy[i];if (dx<1 || dx>n || dy<1 || dy>n || yz[j] != a[dx][dy]) {flag = 0;break;}}if (flag) {for (int j = 0; i < 7; j++) {int dx = x + j * xx[i];int dy = y + j * yy[i];a[dx][dy] = 1;}}}}signed main() {cin >> n;for (int i = 1; i <= n; i++) {for (int j = 1; j <= n; j++) {cin >> a[i][j];}}for (int i = 1; i <= n; i++) {for (int j = 1; j <= n; j++) {if (a[i][j] == 'y')dfs(i, j);}}for (int i = 1; i <= n; i++) {for (int j = 1; j <= n; j++) {if (!vis[i][j])cout << '*';elsecout << a[i][j];}cout << endl;}return 0;
}