字串变换 bfs + 字符串

字串变换 bfs + 字符串

题目描述

已知有两个字串$A,BA,BA,B及一组字串变换的规则（至多666个规则）:$

$A1A_1A1​ ->B1 B_1B1​$

$A2A_2A2​ -> B2B_2B2​$

规则的含义为：在 $AAA中的子串 A1A_1A1​ 可以变换为B1 B_1B1​，A2A_2A2​ 可以变换为 B2B_2B2​ …。$

例如：$AAA=$

变换规则为：

‘$abcabcabc’->‘xuxuxu’‘ududud’->‘yyy’‘yyy’->‘yzyzyz’$

则此时，$AAA可以经过一系列的变换变为BBB，其变换的过程为：$

‘$abcdabcdabcd’->‘xudxudxud’->‘xyxyxy’->‘xyzxyzxyz’$

共进行了$333次变换，使得AAA变换为BBB。$

输入输出格式

输入格式：

输入格式如下：

$AAA BBBA1A_1A1​ B1B_1B1​A2A_2A2​ B2B_2B2​ |-> 变换规则$

... ... /

所有字符串长度的上限为$202020。$

输出格式：

输出至屏幕。格式如下：

若在$101010步（包含101010步）以内能将AAA变换为BBB，则输出最少的变换步数；否则输出"NO ANSWER!"$

输入输出样例

输入样例#1： 复制

abcd xyz abc xu ud y y yz

输出样例#1： 复制

3

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize(2)
using namespace std;
#define maxn 100005
#define inf 0x7fffffff
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long  ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-4
typedef pair<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;
inline ll rd() {
	ll x = 0;
	char c = getchar();
	bool f = false;
	while (!isdigit(c)) {
		if (c == '-') f = true;
		c = getchar();
	}
	while (isdigit(c)) {
		x = (x << 1) + (x << 3) + (c ^ 48);
		c = getchar();
	}
	return f ? -x : x;
}

ll gcd(ll a, ll b) {
	return b == 0 ? a : gcd(b, a%b);
}
int sqr(int x) { return x * x; }


/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
	if (!b) {
		x = 1; y = 0; return a;
	}
	ans = exgcd(b, a%b, x, y);
	ll t = x; x = y; y = t - a / b * y;
	return ans;
}
*/

struct node {
	string str;
	int stp;
};
string a, b;
string orign[maxn];
string trans[maxn];
int n, ans;
map<string, int>mp;
string tran(const string &str, int i, int j) {
	string ans = "";
	if (i + orign[j].length() > str.length())return ans;
	for (int k = 0; k < orign[j].length(); k++) {
		if (str[i + k] != orign[j][k])return ans;
	}
	ans = str.substr(0, i);
	ans += trans[j];
	ans += str.substr(i + orign[j].length());
	return ans;
}
void bfs() {
	queue<node>q;
	node s; s.str = a; s.stp = 0; q.push(s);
	while (!q.empty()) {
		node u = q.front(); q.pop();
		string tmp;
		if (mp.count(u.str))continue;
		if (u.str == b) {
			ans = u.stp; break;
		}
		mp[u.str] = 1;
		for (int i = 0; i < u.str.length(); i++) {
			for (int j = 0; j < n; j++) {
				tmp = tran(u.str, i, j);
				if (tmp != "") {
					node v; v.str = tmp; v.stp = u.stp + 1;
					q.push(v);
				}
			}
		}
	}
	if (ans > 10 || ans == 0)cout << "NO ANSWER!" << endl;
	else cout << ans << endl;
}


int main() {
	ios::sync_with_stdio(0);
	cin >> a >> b;
	while (cin >> orign[n] >> trans[n])n++;
	bfs();
	return 0;
}