2020年3月28日 ICPC训练联盟周赛 比赛补题报告

Posted 2021-03-03 zhengqc

 

 

 

 

 

 

 

 

 

 

 

 

 

 思路：

　　暴力搜索. 建立一个数组alpha，与26个字母相对应. 每个数组包含一个Vector，Vector中保存每个字母的位置.

　　例如alpha[0] 代表存储字母‘A‘ 的Vector;

代码：

　　

  1 #include <iostream>
  2 #include <vector>
  3 #include <string>
  4 #include <cstdio>
  5 using namespace std;
  6 
  7 struct Node {
  8     int x, y;
  9 };
 10 string str[110];
 11 // 数组的数组
 12 vector<Node> alpha[26];
 13 int main() {
 14     int T;
 15     cin >> T;
 16     for (int t = 1; t <= T; ++t) {
 17         for (int j = 0; j < 26; ++j) {
 18             alpha[t].clear();
 19         }
 20         int R, C;
 21         cin >> R >> C;
 22         Node temp;
 23         for (int i = 0; i < R; ++i) {
 24             cin >> str[i];
 25             for (int j = 0; j < C; ++j) {
 26                 temp.x = i, temp.y = j;
 27                 alpha[ str[i][j] - ‘A‘ ].push_back( temp );
 28             }
 29         }
 30 
 31         printf("Word search puzzle #%d:
", t);
 32         int n;
 33         cin >> n;
 34         string query;
 35         for (int i = 0; i < n; ++i) {
 36             cin >> query;
 37             int cnt = alpha[ query[0] - ‘A‘ ].size();
 38 
 39             for (int j = 0; j < cnt; ++j) {
 40                 Node head = alpha[ query[0] - ‘A‘ ][j];
 41                 // 向上查找
 42                 bool flag = true;
 43                 for (int e = (head.x-1+2*R)%R, f = 1; f < query.length(); e = (e-1+2*R)%R, ++f) {
 44                     if (query[f] != str[e][head.y]) {
 45                         flag = false;
 46                         break;
 47                     }
 48                 }
 49                 if (flag) {
 50                     printf("U %d %d ", head.x+1, head.y+1);
 51                     cout << query << endl;
 52                     break;
 53                 }
 54 
 55                 // 向下查找
 56                 flag = true;
 57                 for (int e = (head.x+1)%R, f = 1; f < query.length(); e = (e+1)%R, ++f) {
 58                     if (query[f] != str[e][head.y]) {
 59                         flag = false;
 60                         break;
 61                     }
 62                 }
 63                 if (flag) {
 64                     printf("D %d %d ", head.x+1, head.y+1);
 65                     cout << query << endl;
 66                     break;
 67                 }
 68 
 69                 // 向左查找
 70                 flag = true;
 71                 for (int e = (head.y-1+2*C)%C, f = 1; f < query.length(); e = (e-1+2*C)%C, ++f) {
 72                     if (query[f] != str[head.x][e]) {
 73                         flag = false;
 74                         break;
 75                     }
 76                 }
 77                 if (flag) {
 78                     printf("L %d %d ", head.x+1, head.y+1);
 79                     cout << query << endl;
 80                     break;
 81                 }
 82 
 83                 // 向右查找
 84                 flag = true;
 85                 for (int e = (head.y+1)%C, f = 1; f < query.length(); e = (e+1)%C, ++f) {
 86                     if (query[f] != str[head.x][e]) {
 87                         flag = false;
 88                         break;
 89                     }
 90                 }
 91                 if (flag) {
 92                     printf("R %d %d ", head.x+1, head.y+1);
 93                     cout << query << endl;
 94                     break;
 95                 }
 96             }
 97         }
 98         cout << endl;
 99     }
100 
101     return 0;
102 }