DFS ```cpp #include <iostream> using namespace std; const int N = 1010; char s[] = "tencent"; int n, m; char d[N][N]; bool st[N][N]; int ans; int dx[] = { -1, 0, 1, 0 }; int dy[] = { 0, -1, 0, 1 }; void dfs(int x, int y, int u) { if (u == 6) { ans++; return; } 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 && d[nx][ny] == s[u + 1]) dfs(nx, ny, u + 1); } } void solve() { cin >> n >> m; for (int i = 1; i <= n; i++) for (int j = 1; j <= m; j++) cin >> d[i][j]; for (int i = 1; i <= n; i++) for (int j = 1; j <= m; j++) if (d[i][j] == 't') dfs(i, j, 0); cout << ans << endl; } int main() { solve(); return 0; } ```

``` #include <bits/stdc++.h> using namespace std; const int N=1010; int n,m,ans; char g[N][N]; string target="tencent"; int dx[]={0,0,1,-1}; int dy[]={1,-1,0,0}; void dfs(int x,int y,int pos){ if(pos==6){ ans++; return; } for(int i=0;i<4;i++){ int a=x+dx[i],b=y+dy[i]; if(g[a][b]==target[pos+1]){ dfs(a,b,pos+1); } } } int main(){ cin>>n>>m; for(int i=1;i<=n;i++){ for(int j=1;j<=m;j++){ cin>>g[i][j]; } } for(int i=1;i<=n;i++){ for(int j=1;j<=m;j++){ if(g[i][j]=='t'){ dfs(i,j,0); } } } cout<<ans<<endl; return 0; } //tencent ```

``` #include<bits/stdc++.h> using namespace std; const int N=1010; char s[]="tencent"; int n,m; char d[N][N]; // bool st[N][N]; int ans; int dx[]={-1,0,1,0}; int dy[]={0,-1,0,1}; void dfs(int x,int y,int u) { if(u==6) { ans++; return; } 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&&d[nx][ny]==s[u+1]) { dfs(nx,ny,u+1); } } } int main() { cin>>n>>m; for(int i=1;i<=n;i++) { for(int j=1;j<=m;j++) { cin>>d[i][j]; } } for(int i=1;i<=n;i++) { for(int j=1;j<=m;j++) { if(d[i][j]=='t') { dfs(i,j,0); } } } cout<<ans; return 0; } ```

``` #include <iostream> #include <string> #include <vector> using namespace std; const string target = "tencent"; int n, m; vector<vector<char>> matrix; int directions[4][2] = {{0, 1}, {1, 0}, {-1, 0}, {0, -1}}; int ans = 0; void dfs(int x, int y, int step) { if (step == 6) { ans++; return; } for (int i = 0; i < 4; i++) { int nx = x + directions[i][0]; int ny = y + directions[i][1]; if (nx >= 0 && nx < n && ny >= 0 && ny < m && matrix[nx][ny] == target[step + 1]) { dfs(nx, ny, step + 1); } } } int main() { cin >> n >> m; matrix.resize(n, vector<char>(m)); for (int i = 0; i < n; i++) { for (int j = 0; j < m; j++) { cin >> matrix[i][j]; } } for (int i = 0; i < n; i++) { for (int j = 0; j < m; j++) { if (matrix[i][j] == 't') { dfs(i, j, 0); } } } cout << ans << endl; return 0; } ```

``` 为什么跑样例可以，评测不行呢 #include<bits/stdc++.h> using namespace std; char Map[1010][1010]; int n, m, ans; int d[4][2] = { {-1,0}, {1,0}, {0,-1}, {0,1} }; struct point { int x; int y; int cnt; string str; }; void bfs(int x, int y) { queue<point> que;//重新构造队列 point start; start.x = x, start.y = y, start.cnt = 0; start.str = Map[x][y]; que.push(start); while (!que.empty()) { if (que.front().cnt == 6 && que.front().str == "tencent") { ans++; } else if (que.front().cnt > 6) break; for (int i = 0; i < 4; i++) { int tx = que.front().x + d[i][0], ty = que.front().y + d[i][1]; if (tx >= 1 && tx <= n && ty >= 1 && ty <= m) { point temp; temp.x = tx; temp.y = ty; temp.cnt = que.front().cnt + 1; temp.str = que.front().str + Map[tx][ty]; que.push(temp); } } que.pop(); } } int main() { cin >> n >> m; for (int i = 1; i <= n; i++) for (int j = 1; j <= m; j++) cin >> Map[i][j]; for (int i = 1; i <= n; i++) { for (int j = 1; j <= m; j++) { bfs(i, j); } } cout << ans; return 0; } ```

DFS ``` #include <iostream> #include <cstring> #include <algorithm> using namespace std; const int N = 1e3 + 5; const int INF = 0x3f3f3f3f; char g[N][N]; int vis[N][N]; int n, m ,ans; int dx[] = {0,0,-1,1}; int dy[] = {1,-1,0,0}; string s = "tencent"; void dfs(int x,int y,int c) { if(x < 1 || x > n || y < 1 || y > m) return; if(g[x][y] != s[c]) return; if(c == 6 && g[x][y] == s[6]) { ans++; return; } for(int i = 0;i < 4;i++){ int x1 = x + dx[i];int y1 = y + dy[i]; if(vis[x1][y1]) continue; vis[x1][y1] = 1; dfs(x1,y1,c + 1); vis[x1][y1] = 0; } } int main() { ios::sync_with_stdio(false); cin.tie(0); cout.tie(0); cin >> n >> m; for(int i = 1;i <= n;i++) for(int j = 1;j <= m;j++) cin >> g[i][j]; for(int i = 1;i <= n;i++){ for(int j = 1;j <= m;j++){ if(g[i][j] == 't') dfs(i,j,0); } } cout << ans << endl; return 0; } ```

import java.io.; public class Main { public static BufferedReader in = new BufferedReader(new InputStreamReader(System.in)); public static int n; public static int m; public static String[][] map; public static long cnt = 0; public static boolean[][] isuse; public static String[] t = { "t", "e", "n", "c", "e", "n", "t" }; public static int[][] way = { { -1, 0 }, { 1, 0 }, { 0, -1 }, { 0, 1 } }; ``` public static void main(String[] args) throws IOException { // TODO Auto-generated method stub String[] s = in.readLine().split(" "); n = Integer.parseInt(s[0]); m = Integer.parseInt(s[1]); map = new String[n][m]; isuse = new boolean[n][m]; for (int i = 0; i < n; i++) { map[i] = in.readLine().split(""); } for (int i = 0; i < n; i++) { for (int j = 0; j < m; j++) { if (map[i][j].equals("t")) {//剪枝，从t开始走 dfs(i, j, 1, map[i][j]); } } } System.out.println(cnt); } public static void dfs(int x, int y, int len, String ans) { if (ans.equals("tencent")) { cnt++; return; } if (len > 6)//剪枝 return; for (int i = 0; i < 4; i++) {//枚举四个方向 int xx = x + way[i][0]; int yy = y + way[i][1]; //剪枝，下一步要走的和期望要得到的字符一样才走 if (check(xx, yy) && map[xx][yy].equals(t[len])) { isuse[xx][yy] = true;//走过就不走了 dfs(xx, yy, len + 1, ans + map[xx][yy]); isuse[xx][yy] = false;//回溯 } } } public static boolean check(int x, int y) {//判断是否在边界内，并且可走 return x >= 0 && x < n && y >= 0 && y < m && isuse[x][y] == false; } ``` }

```cpp // https://dashoj.com/p/81 // tencent #include <bits/stdc++.h> #define N 1010 using namespace std; string str = "tencent"; int n, m, ans = 0, dx[4] = {-1, 0, 1, 0}, dy[4] = {0, 1, 0, -1}; char g[N][N]; void dfs(int x, int y, int cnt) { if (cnt == 7) { ans++; return; } char c = str[cnt]; for (int i = 0; i < 4; i++) { int vx = x + dx[i], vy = y + dy[i]; if (vx >= 1 && vx <= n && vy >= 1 && vy <= m && g[vx][vy] == c) dfs(vx, vy, cnt + 1); } } int main() { cin >> n >> m; for (int i = 1; i <= n; i++) for (int j = 1; j <= m; j++) cin >> g[i][j]; for (int i = 1; i <= n; i++) for (int j = 1; j <= m; j++) if (g[i][j] == 't') dfs(i, j, 1); cout << ans << endl; return 0; } ```