题意：自己看题目，中文体面。
题解：
把所有不能走的路径放入AC自动机中。
然后DP【i】【j】表示走到 i 这个点，且位于AC自动机 j 这个节点最短距离
然后直接DP即可。注意一点会爆int
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <cmath>
#include <ctime>
#include <cstdio>
#include <string>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <unordered_map> #define pi acos(-1.0)
#define eps 1e-9
#define fi first
#define se second
#define rtl rt<<1
#define rtr rt<<1|1
#define bug printf("******\n")
#define mem(a, b) memset(a,b,sizeof(a))
#define name2str(x) #x
#define fuck(x) cout<<#x" = "<<x<<endl
#define sfi(a) scanf("%d", &a)
#define sffi(a, b) scanf("%d %d", &a, &b)
#define sfffi(a, b, c) scanf("%d %d %d", &a, &b, &c)
#define sffffi(a, b, c, d) scanf("%d %d %d %d", &a, &b, &c, &d)
#define sfL(a) scanf("%lld", &a)
#define sffL(a, b) scanf("%lld %lld", &a, &b)
#define sfffL(a, b, c) scanf("%lld %lld %lld", &a, &b, &c)
#define sffffL(a, b, c, d) scanf("%lld %lld %lld %lld", &a, &b, &c, &d)
#define sfs(a) scanf("%s", a)
#define sffs(a, b) scanf("%s %s", a, b)
#define sfffs(a, b, c) scanf("%s %s %s", a, b, c)
#define sffffs(a, b, c, d) scanf("%s %s %s %s", a, b,c, d)
#define FIN freopen("../in.txt","r",stdin)
#define gcd(a, b) __gcd(a,b)
#define lowbit(x) x&-x
#define IO iOS::sync_with_stdio(false) using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
const ULL seed = ;
const LL INFLL = 0x3f3f3f3f3f3f3f3fLL;
const int maxn = 1e6 + ;
const int maxm = 8e6 + ;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + ;
int n, m, k, x[maxn], y[maxn], buf[]; double cal(int id1, int id2) {
return sqrt((1.0 * x[id1] - x[id2]) * (1.0 * x[id1] - x[id2]) + (1.0 * y[id1] - y[id2]) * (1.0 * y[id1] - y[id2]));
} struct Aho_Corasick {
int next[][], fail[], End[];
int root, cnt; int newnode() {
for (int i = ; i < ; i++) next[cnt][i] = -;
End[cnt++] = ;
return cnt - ;
} void init() {
cnt = ;
root = newnode();
} void insert(int len) {
int now = root;
for (int i = ; i < len; i++) {
if (next[now][buf[i]] == -) next[now][buf[i]] = newnode();
now = next[now][buf[i]];
}
End[now]++;
} void build() {
queue<int> Q;
fail[root] = root;
for (int i = ; i < ; i++)
if (next[root][i] == -) next[root][i] = root;
else {
fail[next[root][i]] = root;
Q.push(next[root][i]);
}
while (!Q.empty()) {
int now = Q.front();
Q.pop();
if (End[fail[now]]) End[now] = ;
for (int i = ; i < ; i++)
if (next[now][i] == -) next[now][i] = next[fail[now]][i];
else {
fail[next[now][i]] = next[fail[now]][i];
Q.push(next[now][i]);
}
}
} double dp[][]; void solve() {
for (int i = ; i <= n; ++i)
for (int j = ; j < cnt; ++j)
dp[i][j] = INFLL;
dp[][next[root][]] = ;
for (int i = ; i <= n; ++i) {
for (int j = ; j < cnt; ++j) {
if (dp[i][j] == INFLL) continue;
for (int k = i + ; k <= n; ++k) {
int idx = next[j][k];
if (End[idx]) continue;
dp[k][idx] = min(dp[k][idx], dp[i][j] + cal(i, k));
}
}
}
double ans = INFLL;
for (int i = ; i < cnt; ++i) ans = min(ans, dp[n][i]);
if (ans == INFLL) printf("Can not be reached!\n");
else printf("%.2f\n", ans);
} void debug() {
for (int i = ; i < cnt; i++) {
printf("id = %3d,fail = %3d,end = %3d,chi = [", i, fail[i], End[i]);
for (int j = ; j < ; j++) printf("%2d", next[i][j]);
printf("]\n");
}
}
} ac; int main() {
// FIN;
while (~sffi(n, m)) {
if (n == && m == ) break;
for (int i = ; i <= n; ++i) sffi(x[i], y[i]);
ac.init();
for (int i = ; i < m; ++i) {
sfi(k);
for (int j = ; j < k; ++j) sfi(buf[j]);
ac.insert(k);
}
ac.build();
ac.solve();
}
return ;
}