HDU5487 Difference of Languages(BFS)

题意：给你两个自动机，求出最短的（如果有相同最短的则求出字典序最小的）能被其中一个自动机接收而不能被另外一个自动机接收的字符串。

一看是自动机以为是神题，后来比赛最后才有思路。

两个自动机的状态都是小于1000的，所以可以建一个图，每个结点（u,v）表示当前处于自动机1的状态u和自动机2的状态v，然后相应的这些状态接收[a-z]的字符就会转移到下一个状态。然后从原点（0,0）开始广搜，搜到的第一个accpet[u]!=accept[v]的即是所求的状态。（处理的时候要给每个自动机加一个状态，用来表示失配的时候的情况，这个状态的所有后继都转移向自己，而且本身不是accpet状态，广搜即可）

#pragma warning(disable:4996)
#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <vector>
#include <cmath>
#include <queue>
using namespace std;

#define maxn 1005

int n1, m1, k1;
int n2, m2, k2;

bool tar1[maxn];
bool tar2[maxn];

int go1[maxn][26];
int go2[maxn][26];

int prestate[maxn*maxn];
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, &m1, &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 < 26; ++k){
if (go1[i][k] == -1) go1[i][k] = n1;
}
}

scanf("%d%d%d", &n2, &m2, &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 < 26; ++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 < 26; ++k){
int nstate = nextState(state, k);
if (!vis[nstate]){
Q.push(nstate);
vis[nstate] = true;
prestate[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());
printf("Case #%d: %s\n", ++ca, ts.c_str());
}
return 0;
}