想到的解法比较rz，就是一层层地进入罐头，当遇到 ’ { ‘，’ [ ‘，’ ( ’ 时进入下一层罐头，当遇到 ’ } ‘，’ ] ‘，’ ) '时检查与罐头开头是否相匹配，若不匹配则彻底false，反之，继续检查。

代码：

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#include<iostream>
#include<string>
#include<map>
using namespace std;
const int maxn = 1005;
string s;
int now = 0;		//检查到下标为now的点
map<char, char>mp;   //对应情况
bool ok = true;		//是否正确
void solve(int len,int depth) {
	char st = s[now];
	now++;
	while(now<len){
		if (s[now] == '}' || s[now] == ']' || s[now] == ')') {
			if (s[now] != mp[st]) {
				ok = false;		//彻底false
			}
			return;
		}
		if (s[now] == '{' || s[now] == '[' || s[now] == '(') {
			solve(len, depth + 1);
		}
		now++;
	}
}
int main() {
	mp['{']='}';
	mp['['] = ']';
	mp['('] = ')';
	int N;
	cin >> N;
	for(int i=1;i<=N;i++) {
		ok = true;
		now = 0;
		cin >> s;
		if (s[0] != '{'&&s[0] != '['&&s[0] != '(') {//开头不是罐头，直接false
			cout << "NO" << endl;
			continue;
		}
		int len = s.length();
		solve(len, 0);
		if (ok && now == len - 1)	//ok正确且最后一个点是罐头盖子
			cout << "YES" << endl;
		else cout << "NO" << endl;
	}
	return 0;
}