UVA - 310 L--system（bfs+hash）
| L-system |
A D0L (Deterministic Lindenmayer system without interaction) system consists of a finite set of symbols (the alphabet), a finite set P of productions and a starting string . The productions in P are of the form , where and (u is called the right side of the production), is the set of all strings of symbols from excluding the empty string. Such productions represent the transformation of the symbol x into the string u. For each symbol , P contains exactly one production of the form . Direct derivation from string to consists of replacing each occurrence of the symbol in by the string on the right side of the production for that symbol. The language of the D0L system consists of all strings which can be derived from the starting string by a sequence of the direct derivations.
Suppose that the alphabet consists of two symbols a and b. So the set of productions includes two productions of the form a , b , where u and , and the starting string . Can you answer whether there exists a string in the language of the D0L system of the form xzy for a given string z? (x and y are some strings from , is the set of all strings of symbols from , including the empty string.). Certainly you can. Write the program which will solve this problem.
Input
The input file of the program consists of several blocks of lines. Each block includes four lines. There are no empty lines between any successive two blocks. The first line of a block contains the right side of the production for the symbol a. The second one contains the right side of the production for the symbol band the third one contains the starting string and the fourth line the given string z. The right sides of the productions, the given string z and the starting string are at most 15 characters long.
Output
For each block in the input file there is one line in the output file containing YES or NO according to the solution of the given problem.
Sample Input
aa bb ab aaabb a b ab ba
Sample Output
YES NO
这题实在是难以理解，并且在写的时候，把一个步骤写成死循环了，一直TLE。
题意：
给出4个字符串
第一个是字符'a'的替换字符串，
第二个是字符'b'的替换字符串，
第三个是初始字符串，
第四个是目标字符串。
问是否初始字符串是否可以经过"L-system"替换形成目标字符串。
替换规则：
(1)将初始的字符串的a和b进行替换，
(2)再将新串中和 目标串长度的子串 提取出来，判断是否相同，如果相同返回"YES"
(3)如果不相同，将子串中的a和b进行替换，并跳到步骤(2)
解析：这题用到bfs+hash，bfs用于枚举出所有的子串，而hash用于判断是否重复。
#include <cstdio>
#include <cstring>
#include <string>
#include <set>
#include <queue>
#include <iostream>
using namespace std;
const int MAX = 1 << 16;
string a,b,begin,end;
queue<string> q;
bool vis[MAX];
int hash(string str) {
int v = 0;
for(int i = 0; i < str.size(); i++) {
v = v*2 + str[i] - 'a' + 1;
}
return v;
}
void init() {
memset(vis,0,sizeof(vis));
while(!q.empty()) {
q.pop();
}
}
bool bfs() {
init();
string tmp;
int v,over = hash(end);
for(int i = 0; i < begin.size(); i++) {
tmp = "";
for(int j = i,k = 0; j < begin.size() && k < end.size(); j++,k++) {
tmp += begin[j];
}
v = hash(tmp);
if(v == over) {
return true;
}else if(!vis[v]) {
vis[v] = true;
q.push(tmp);
}
}
while(!q.empty()) {
string front = q.front();
q.pop();
tmp = "";
for(int i = 0; i < front.size(); i++) {
if(front[i] == 'a') {
tmp += a;
}else {
tmp += b;
}
}
for(int i = 0; i < tmp.size(); i++) {
string tmp2 = "";
for(int j = i,k = 0; j < tmp.size() && k < end.size(); j++,k++) {
tmp2 += tmp[j];
}
v = hash(tmp2);
if(v == over) {
return true;
}else if(!vis[v]) {
vis[v] = true;
q.push(tmp2);
}
}
}
return false;
}
int main() {
bool ok;
while(cin >> a) {
cin >> b >> begin >> end;
ok = bfs();
if(ok) {
cout << "YES" << endl;
}else {
cout << "NO" <<endl;
}
}
return 0;
}