小明系列故事――女友的考验 HDU

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("******
")
#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 = 13331;
const LL INFLL = 0x3f3f3f3f3f3f3f3fLL;
const int maxn = 1e6 + 7;
const int maxm = 8e6 + 10;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
int n, m, k, x[maxn], y[maxn], buf[105];

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[510][52], fail[510], End[510];
    int root, cnt;

    int newnode() {
        for (int i = 0; i < 52; i++) next[cnt][i] = -1;
        End[cnt++] = 0;
        return cnt - 1;
    }

    void init() {
        cnt = 0;
        root = newnode();
    }

    void insert(int len) {
        int now = root;
        for (int i = 0; i < len; i++) {
            if (next[now][buf[i]] == -1) next[now][buf[i]] = newnode();
            now = next[now][buf[i]];
        }
        End[now]++;
    }

    void build() {
        queue<int> Q;
        fail[root] = root;
        for (int i = 0; i < 52; i++)
            if (next[root][i] == -1) 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] = 1;
            for (int i = 0; i < 52; i++)
                if (next[now][i] == -1) next[now][i] = next[fail[now]][i];
                else {
                    fail[next[now][i]] = next[fail[now]][i];
                    Q.push(next[now][i]);
                }
        }
    }

    double dp[52][510];

    void solve() {
        for (int i = 0; i <= n; ++i)
            for (int j = 0; j < cnt; ++j)
                dp[i][j] = INFLL;
        dp[1][next[root][1]] = 0;
        for (int i = 1; i <= n; ++i) {
            for (int j = 0; j < cnt; ++j) {
                if (dp[i][j] == INFLL) continue;
                for (int k = i + 1; 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 = 0; i < cnt; ++i) ans = min(ans, dp[n][i]);
        if (ans == INFLL) printf("Can not be reached!
");
        else printf("%.2f
", ans);
    }

    void debug() {
        for (int i = 0; i < cnt; i++) {
            printf("id = %3d,fail = %3d,end = %3d,chi = [", i, fail[i], End[i]);
            for (int j = 0; j < 26; j++) printf("%2d", next[i][j]);
            printf("]
");
        }
    }
} ac;

int main() {
  //  FIN;
    while (~sffi(n, m)) {
        if (n == 0 && m == 0) break;
        for (int i = 1; i <= n; ++i) sffi(x[i], y[i]);
        ac.init();
        for (int i = 0; i < m; ++i) {
            sfi(k);
            for (int j = 0; j < k; ++j) sfi(buf[j]);
            ac.insert(k);
        }
        ac.build();
        ac.solve();
    }
    return 0;
}