标签：cto show for print algo 方向 goto 避免 返回
oj地址是韩国网站 连接比较慢 https://algospot.com/judge/problem/read/BOGGLE
大意如下
输入输出
输入 1 URLPM XPRET GIAET XTNZY XOQRS 6 PRETTY GIRL REPEAT KARA PANDORA GIAZAPX 输出 PRETTY YES GIRL YES REPEAT YES KARA NO PANDORA NO GIAZAPX YES
估摸着很简单 就蹭蹭8个方向DFS 代码写完
测试用例过了
代码如下
#include <iostream> #include <string> #include <vector> using namespace std; int n, m; int record = 0; const int dx[8] = { -1,-1,-1,1,1,1,0,0 }; const int dy[8] = { -1,0,1,-1,0,1,-1,1 }; bool isrange(int x, int y) { if (x < 0 || x >= 5 || y < 0 || y >= 5) return false; return true; } bool hasword(int x, int y, const string& word ,int idx,const vector<vector<char>>& table) { if (!isrange(x, y)) return false; if (table[x][y] != word[idx]) return false; if (idx == word.size()-1) return true; for (int i = 0; i < 8; i++) { int nextx = x + dx[i]; int nexty = y + dy[i]; if (hasword(nextx, nexty, word,idx+1, table)) return true; } return false; } int main() { int t = 0; cin >> t; while (t--) { vector<vector<char>> table(5, vector<char>(5, 0)); for (int i = 0; i < 5; i++) { for (int j = 0; j < 5; j++) { cin >> table[i][j]; } } int check = 0; cin >> check; vector<string> checkWord; while (check--) { string s; cin >> s; int find = 0; record = 0; for (int i = 0; i < 5; i++) { for (int j = 0; j < 5; j++) { if (hasword(i, j, s, 0,table)) { cout << s << " YES" << endl; find = 1; goto LABEL; } } } LABEL: if (0 == find) { cout << s << " NO" << endl; } } } return 0; }
但是居然一个特殊用例过不了 代码被判为TLE
输入 1 AAAAA AAAAA AAAAA AACCC AACCB 1 AAAAAAAAAB
左思右想不得其门 只得针对该例子进行了特殊处理 翻转字符串！
添加代码不多 仅仅多了 reverse(s.begin(), s.end()); 一行
1 #include <iostream> 2 3 #include <iostream> 4 #include <string> 5 #include <vector> 6 #include <algorithm> 7 8 using namespace std; 9 10 int n, m; 11 12 const int dx[8] = { -1,-1,-1,1,1,1,0,0 }; 13 const int dy[8] = { -1,0,1,-1,0,1,-1,1 }; 14 15 vector<vector<int>> mem(5, vector<int>(5, 0)); 16 17 bool isrange(int x, int y) { 18 if (x < 0 || x >= 5 || y < 0 || y >= 5) 19 return false; 20 21 return true; 22 } 23 24 25 26 bool hasword(int x, int y, const string& word, int idx, const vector<vector<char>>& table) 27 { 28 if (!isrange(x, y)) { 29 return false; 30 } 31 32 if (table[x][y] != word[idx]) { 33 34 return false; 35 } 36 37 if (idx >= word.size() - 1) return true; 38 39 for (int i = 0; i < 8; i++) { 40 int nextx = x + dx[i]; 41 int nexty = y + dy[i]; 42 if (hasword(nextx, nexty, word, idx + 1, table)) 43 return true; 44 } 45 46 return false; 47 } 48 49 50 51 int main() 52 { 53 int t = 0; 54 cin >> t; 55 while (t--) { 56 vector<vector<char>> table(5, vector<char>(5, 0)); 57 for (int i = 0; i < 5; i++) { 58 for (int j = 0; j < 5; j++) { 59 cin >> table[i][j]; 60 } 61 } 62 63 int check = 0; 64 cin >> check; 65 vector<string> checkWord; 66 67 while (check--) { 68 string s; 69 cin >> s; 70 string copys = s; 71 reverse(s.begin(), s.end()); 72 int find = 0; 73 for (int i = 0; i < 5; i++) { 74 for (int j = 0; j < 5; j++) { 75 mem[i][j] = 1; 76 } 77 } 78 for (int i = 0; i < 5; i++) { 79 for (int j = 0; j < 5; j++) { 80 if (hasword(i, j, s, 0, table)) { 81 cout << copys << " YES" << endl; 82 find = 1; 83 goto LABEL; 84 } 85 } 86 } 87 88 LABEL: 89 if (0 == find) { 90 cout << copys << " NO" << endl; 91 } 92 } 93 } 94 95 96 return 0; 97 }
但是实际上 当测试字符串为AAAAAAAAABAAAAAAAA 也一样会TLE
所以 实际上我们需要使用数组缓存从当前格子出发不能解决的字符串记录 避免多次重复尝试失败的路径
代码添加了一个变量数组
int d[x][y][l]; 记录x y格子出发的尝试匹配长度为l的字符串 能否成功，如果失败则置0。
下次DFS 发现该标记为 0 则直接返回 不进行尝试
1 #include <bits/stdc++.h> 2 using namespace std; 3 int T,n,c[15][15],cw[15],len[15]; 4 char a[15][15],w[15][15]; 5 int nx[10] = {1,0,-1,-1,-1,0,1,1},ny[10] = {1,1,1,0,-1,-1,-1,0}; 6 int dx,dy,is; 7 int d[15][15][15]; 8 9 int dfs(int x,int y,int now,int l) 10 { 11 if(d[x][y][l]) return 0; 12 if(l == len[now]) return 1; 13 d[x][y][l] = 1; 14 is = 0; 15 for(int i = 0;i < 8;i++) 16 { 17 dx = x+nx[i]; 18 dy = y+ny[i]; 19 if(a[dx][dy] == w[now][l]) 20 { 21 if(dfs(dx,dy,now,l+1)) 22 { 23 //c[x][y] = 0; 24 return 1; 25 } 26 } 27 } 28 //d[x][y][l] = 0; 29 return 0; 30 } 31 32 int main() 33 { 34 scanf("%d",&T); 35 while(T--) 36 { 37 memset(cw,0,sizeof(cw)); 38 for(int i = 1;i <= 5;i++) 39 { 40 for(int j = 1;j <= 5;j++) 41 { 42 scanf(" %c",&a[i][j]); 43 } 44 } 45 scanf("%d",&n); 46 for(int i = 1;i <= n;i++) 47 { 48 scanf("%s",w[i]); 49 len[i] = strlen(w[i]); 50 for(int j = 1;j <= 5;j++) 51 { 52 for(int k = 1;k <= 5;k++) 53 { 54 if(a[j][k] == w[i][0]) 55 { 56 if(dfs(j,k,i,1)) 57 { 58 cw[i] = 1; 59 break; 60 } 61 } 62 } 63 if(cw[i]) break; 64 } 65 memset(d,0,sizeof(d)); 66 } 67 for(int i = 1;i <= n;i++) 68 { 69 printf("%s %s\n",w[i],((cw[i] == 1) ? "YES" : "NO")); 70 } 71 } 72 }
标签：cto show for print algo 方向 goto 避免 返回
原文地址：https://www.cnblogs.com/itdef/p/11524930.html
