第二个测试样例应该输出8，但是答案是9，是不是有问题啊 附上第二个测试点数据:

((T+F)+(F+F))+(F+T) F T+T F T+F T F+T T F+F F (T) T (F) F

题目给的输出是9，我觉得8次也行

下面是我跑出来的

F
T+T
(T)+T
(T+F)+T
(T+(F))+T
((T)+(F))+(T)
((T)+(F+F))+(T)
((T)+(F+F))+(F+T)
((T+F)+(F+F))+(F+T)

#include <iostream>
#include <string>
#include <cstring>
#include <algorithm>
#include <map>
using namespace std;
const int N = 1e6;
string a, b;
struct node{
	string s1, s2;
}s[10];
int cnt;
struct node1{
	string str;
	int step, f;
}deq[N];
int front, tail, ans, tag;
map<string, int> Maps;

int main()
{
	cin>>a>>b;
	while (cin>>s[cnt].s1>>s[cnt].s2) cnt++;
	deq[tail].str = a; deq[tail].step = 0;deq[tail].f = -1;
	tail++; Maps[a] = 0;
	while (front < tail){
		node1 tn = deq[front++];
		if (tn.step >= 10) continue;
		for (int i = 0; i < cnt; i++){
			if (tn.str.find(s[i].s1) == string::npos) continue;
			string ts = tn.str;
			int index = 0;
			while (true){
				index = ts.find(s[i].s1, index);
				if (index == string::npos) break;
				ts.replace(index, s[i].s1.length(), s[i].s2);
				index += s[i].s1.length();
//				cout<<"ts == "<<ts<<endl;
			}
			if ((Maps.count(ts) && Maps[ts] > tn.step + 1) || !Maps.count(ts)){
				Maps[ts] = deq[tail].step = tn.step + 1;
				deq[tail].str = ts; deq[tail].f = front - 1;
				tail++;
			}
			if (ts == b){
				tag = 1;
				ans = deq[tail - 1].step;
				break;
			}
		}
		if (tag) break;
	}
	if (tag) cout<<ans<<endl;
	else cout<<"NO ANSWER!"<<endl;
	int move = tail - 1;
	while (move != -1){
		cout<<deq[move].str<<endl;
		move = deq[move].f;
	}
	
	
	
	return 0;
}