【HDU 4511】小明系列故事——女友的考验（AC自动机+DP）

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<queue>
#include<vector>
#include<cmath>
using namespace std;
const int     MAXN = 6000;
const int     MAXM = 600;
const double  INF  = 1e18;
int     trie[600][60],cnt;
int     fail[600],val[600],n,m,k;
double  dp[60][600];
struct node
{
    double x, y;
}a[60];
void insert(double *v)
{
    int now = 0;
    for (int i = 1; i <=k; i++)
    {
        int u = v[i];
        if (!trie[now][u])
        {
            trie[now][u] = ++cnt;
        }
        now = trie[now][u];
    }
    val[now]=1;
}
void get_fail()
{
    int now = 0;
    queue<int> q;
    for (int i = 1; i <=n; i++)
        if (trie[now][i])
        {
            fail[trie[now][i]] = 0;
            q.push(trie[now][i]);
        }
    while (!q.empty())
    {
        int u = q.front(); q.pop();
        if (val[fail[u]])
            val[u]=1;
        for (int i = 1; i <=n; i++)
        {
            int v = trie[u][i];
            if (v)
            {
                fail[v] = trie[fail[u]][i];
                q.push(v);
            }
            else
            {
                trie[u][i] = trie[fail[u]][i];
            }
        }
    }
    return;
}
double dis(int i, int j)
{
    double tmp = (a[i].x - a[j].x)*(a[i].x - a[j].x) + (a[i].y - a[j].y)*(a[i].y - a[j].y);
    return sqrt(tmp);
}
double v[600];
double solve()
{
    double minn = INF;
    for (int i = 1; i <= n; i++)
        for (int j = 0; j <= cnt; j++)
            dp[i][j] = INF;
    dp[1][trie[0][1]] = 0;
    for (int i = 1; i <= n; i++)
    {
        for (int j = 0; j <= cnt; j++)
        {
            if (dp[i][j]<INF)
            {
                for (int k = i + 1; k <= n; k++)
                    if (!val[trie[j][k]])
                        dp[k][trie[j][k]] = min(dp[k][trie[j][k]], dp[i][j] + dis(i, k));
            }
        }
    }
    for(int i=0;i<=cnt;i++)
            minn = min(minn, dp[n][i]);
    return minn;
}
int main()
{
    while (scanf("%d%d", &n, &m) && n && m)
    {
        cnt = 0;
        memset(trie, 0, sizeof(trie));
        memset(val, 0, sizeof(val));
        memset(fail, 0, sizeof(fail));
        for (int i = 1; i <= n; i++)
            scanf("%lf%lf", &a[i].x, &a[i].y);
        while(m--)
        {

            scanf("%d", &k);
            for (int i = 1; i <= k; i++)
                scanf("%lf", &v[i]);
            insert(v);
        }
        get_fail();
        double ans = solve();
        if (ans >= INF)
            printf("Can not be reached!\n");
        else
            printf("%.2f\n", ans);
    }
    return 0;
}