字串变换 bfs + 字符串-CFANZ编程社区

题目描述

已知有两个字串字串变换 bfs + 字符串_#include

字串变换 bfs + 字符串_git_02

字串变换 bfs + 字符串_#define_03

规则的含义为：在字串变换 bfs + 字符串_#include_04

例如：字串变换 bfs + 字符串_#define_05

变换规则为：

‘ 字串变换 bfs + 字符串_#include_06

则此时，字串变换 bfs + 字符串_#define_07

‘ 字串变换 bfs + 字符串_#define_08

共进行了字串变换 bfs + 字符串_#define_09

输入输出格式

输入格式：

输入格式如下：

字串变换 bfs + 字符串_#include_10

... ... /

所有字符串长度的上限为字串变换 bfs + 字符串_git_11

输出格式：

输出至屏幕。格式如下：

若在字串变换 bfs + 字符串_git_12

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

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
//#include
//#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 pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair 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;
mapmp;
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() {
  queueq;
  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;
}

EPFL - Fighting