HDU5487 difference of Languages (BFS)

Test instructions: Give you two automata, find the shortest (if there is the same shortest dictionary order of the smallest) can be received by one of the automata and can not be received by another automaton machine string.

A look is the automaton thought is divine question, later race finally had the thought.

Two automata state is less than 1000, so you can build a diagram, each node (U,V) represents the current state of the Automaton 1 and the state of the Automaton 2 V, and then the corresponding states receive [A-z] character will be transferred to the next state. Then from the origin (0,0) began wide search, found the first accpet[u]!=accept[v] is the state of the request. (When processing the time to give each automaton a state, used to indicate the mismatch when the situation, the state of all subsequent transfer to themselves, and not itself Accpet State, wide search can be)

#pragma warning (disable:4996) #include <iostream> #include <cstring> #include <cstdio> #include < algorithm> #include <vector> #include <cmath> #include <queue>using namespace std; #define MAXN 1005int N1, M1, K1;int n2, M2, k2;bool tar1[maxn];bool tar2[maxn];int go1[maxn][26];int go2[maxn][26];int prestate[maxn*m Axn];char prechar[maxn*maxn];bool vis[maxn*maxn];bool Check (int x) {return tar1[x/n2]! = tar2[x%n2];} int nextstate (int x, int c) {return go1[x/n2][c] * n2 + go2[x%n2][c];} int main () {int T; cin >> T; int ca = 0;while (t--) {memset (Go1,-1, sizeof (GO1)), memset (Go2,-1, sizeof (GO2)); Memset ( TAR1, 0, sizeof (TAR1)), memset (tar2, 0, sizeof (TAR2)), int accept;int UI, Vi;char ci[3];scanf ("%d%d%d", &n1, &AMP;M1, & AMP;K1); for (int i = 0; i < K1; ++i) {scanf ("%d", &accept); tar1[accept] = true;} for (int i = 0; I < M1; ++i) {scanf ("%d%d%s", &ui, &vi, CI); go1[ui][ci[0]-' a '] = VI;} for (int i = 0, I <= N1; ++i) {for (int k = 0; K &LT 26; ++k) {if (go1[i][k] = =-1) go1[i][k] = n1;}} scanf ("%d%d%d", &n2, &m2, &AMP;K2); for (int i = 0; i < K2; ++i) {scanf ("%d", &accept); tar2[accept] = true;} for (int i = 0; i < m2; ++i) {scanf ("%d%d%s", &ui, &vi, CI); go2[ui][ci[0]-' a '] = VI;} for (int i = 0, I <= n2; ++i) {for (int k = 0; k <; ++k) {if (go2[i][k] = = 1) go2[i][k] = n2;}} ++N1; ++n2;int ans = -1;memset (Vis, 0, sizeof (VIS));queue<int> Q; Q.push (0); Vis[0] = True;while (! Q.empty ()) {int state = Q.front (); Q.pop (); if (check (state)) {ans = state;break;} for (int k = 0; K <: ++k) {int nstate = Nextstate (state, k); if (!vis[nstate]) {Q.push (nstate); Vis[nstate] = True;prest Ate[nstate] = state;prechar[nstate] = k;}}} if ( -1 = = ans) {printf ("Case #%d:0\n", ++ca); continue;} String Ts;while (ans! = 0) {ts.push_back (char (' a ' + Prechar[ans])); ans = Prestate[ans];} Reverse (Ts.begin (), Ts.end ());p rintf ("Case #%d:%s\n", ++ca, Ts.c_str ());} return 0;}

HDU5487 difference of Languages (BFS)