1. 题目链接

Gift Carpet

2. 题目代码

正确代码

#include<iostream>
#include<vector>
using namespace std;
char letterTable[21][21];
int main(){
    int testCase;
    cin >> testCase;
    int row;
    int column;
    string resultOfCheck;
    while(testCase --){
        cin >> row >> column;
        for(int subscriptOfRow = 1; subscriptOfRow <= row; subscriptOfRow ++){
            for(int subscriptOfColumn = 1; subscriptOfColumn <= column; subscriptOfColumn ++){
                cin >> letterTable[subscriptOfRow][subscriptOfColumn];
            }
        }
        for(int subscriptOfColumn = 1; subscriptOfColumn <= column; subscriptOfColumn ++){
            for(int subscriptOfRow = 1; subscriptOfRow <= row; subscriptOfRow ++){
                if(letterTable[subscriptOfRow][subscriptOfColumn] == 'v' && resultOfCheck.find('v') == string::npos){
                    resultOfCheck = resultOfCheck + 'v';
                    break;
                }else if(letterTable[subscriptOfRow][subscriptOfColumn] == 'i' && resultOfCheck.find('i') == string::npos && resultOfCheck.find('v') != string::npos){
                    resultOfCheck = resultOfCheck + 'i';
                    break;
                }else if(letterTable[subscriptOfRow][subscriptOfColumn] == 'k'&& resultOfCheck.find('k') == string::npos && resultOfCheck.find('i') != string::npos && resultOfCheck.find('v') != string::npos){
                    resultOfCheck = resultOfCheck + 'k';
                    break;
                }else if(letterTable[subscriptOfRow][subscriptOfColumn] == 'a'&& resultOfCheck.find('a') == string::npos && resultOfCheck.find('k') != string::npos && resultOfCheck.find('i') != string::npos && resultOfCheck.find('v') != string::npos){
                    resultOfCheck = resultOfCheck + 'a';
                    break;
                }
            }
        }
        //cout << resultOfCheck << endl;
        if(resultOfCheck == "vika"){
            cout << "YES" << endl;
        }else{
            cout << "NO" << endl;
        }
        resultOfCheck = "";
    }
    return 0;
}

3. 题目总结

解题时长：1h30min左右
保存"vika"相关字符：
① 最初的思路：使用vector容器，解决去重问题有点麻烦(想过使用find函数，但是没有解决)。
② 中间的思路：使用set容器，弊端在于set容器内的字符不是按插入顺序保存的，难以判断字符是否是按"vika"的顺序
③ 最终的思路：使用字符串保存
注意：一列只能出一个字符

学习

#include <bits/stdc++.h>
#define int long long
using namespace std;
 
int t, n, m, b, cnt;
string str[21], a = "vika";
 
signed main() {
	ios::sync_with_stdio(0);
	cin.tie(0), cout.tie(0);
	cin >> t;
	while (t--) {
		cnt = b = 0;
		cin >> n >> m;
		for (int i = 0; i < n; i++) {
			cin >> str[i];
		}
		for (int i = 0; i < m; i++)
			for (int j = 0; j < n; j++) 
				if (str[j][i] == a[b]) {
					b++, cnt++;
					break;
				}
		if (cnt == 4) cout << "yes\n";
		else cout << "no\n";
	}
	return 0;
}