AC automata + two-dimensional shortest path HDU 4511 Xiao Ming series Story--the test of the girlfriend

This question is still relatively good to think of.

Start by setting up an AC automaton for all the non-possible schemes, and then run the shortest way.

First put xiaoming at (sta = 0,pos = 0), STA represents the number of points on the AC automaton, POS represents the number of coordinate points.

According to the POS enumeration where the next can be reached [pos+1,n], then STA moves on the automaton, and if a step causes the STA to be located in a labeled node, then this step is not possible.

#include <iostream> #include <time.h> #include <stdio.h> #include <string.h> #include < stdlib.h> #include <string> #include <map> #include <vector> #include <algorithm> #include <queue> #include <cmath> #define LL long long#define ULL unsigned long longusing namespace Std;const int maxs =    55,MAXN = 510;const int matn = 70;struct mat{int row,col;    ULL Mat[matn][matn];        void Init (int r,int c,int val) {row = R,col = C; for (int i = 1;i <= row, ++i) for (int j = 1;j <= col; ++j) mat[i][j] = (i = = J val:    0);        } Mat Multi (Mat c,ll MOD) {mat tmp; Tmp.        Init (this->row,c.col,0);       int i,j,k; for (k = 1;k <= this->col; ++k) for (i = 1;i <= tmp.row; ++i) for (j = 1;j <= tmp.col;        ++J) Tmp.mat[i][j] + = (this->mat[i][k]*c.mat[k][j]);    return TMP; } Mat Quick (LL n,ll MOD) {matRes,tmp = *this; Res.        Init (row,col,1); while (n) {if (n&1) res = res.            Multi (TMP,MOD); TMP = tmp.            Multi (TMP,MOD);        n >>= 1;    } return res;        } void Output () {cout<< "****************" <<endl;        int i,j;             for (i = 1;i <= row; ++i) {for (j = 1;j <= col; ++j) printf ("%3lld", Mat[i][j]);        Puts ("");    } cout<< "&&&&&&&&&&&&&" <<endl;    }};struct trie{int NEXT[MAXS];    int fail; int flag;} st[maxn];queue<int> q;struct pos{int x, y;}    pos[55];struct acautomaton{int top,root;        int creat () {memset (st[top].next,-1,sizeof (St[top].next));        St[top].flag = 0;        St[top].fail =-1;    return top++;        } void Init () {Top = 0;    root = creat (); } Inline INT Calindex (int c) {return C;        } void Insert (int *s) {int i = 0,TR = root,tmp;            while (s[i]! =-1) {tmp = Calindex (s[i]);            if (st[tr].next[tmp] = =-1) st[tr].next[tmp] = creat ();        tr = st[tr].next[tmp],++i;    } St[tr].flag = 1;        } void Getfail () {st[root].fail =-1;        Q.push (root);        int f,t;            while (q.empty () = = False) {f = Q.front ();            Q.pop ();  for (int i = 0; i < Maxs; ++i) {if (st[f].next[i]! =-1) {T                    = St[f].fail;                    while (t! =-1 && st[t].next[i] = =-1) t = st[t].fail;                    if (t = =-1) st[st[f].next[i]].fail = root;                    else St[st[f].next[i]].fail = St[t].next[i];                Q.push (St[f].next[i]);            }}}} int Match (char *s) {int I, tr = root,tmp;        int ans = 0; for (i = 0; s[i]! = ' + '; ++i) {if (S[i] < ' A ' | |                ' Z ' < s[i]) {tr = root;            Continue            } tmp = Calindex (s[i]);            while (tr! =-1 && st[tr].next[tmp] = =-1) tr = st[tr].fail;                if (tr = =-1) {tr = root;            Continue            } tr = st[tr].next[tmp];            TMP = TR; while (tmp! = root && St[tmp].flag! =-1) {if (St[tmp].flag) ans++,st[            Tmp].flag =-1;    }} return ans; }};int num[10];d ouble dis[510][55];d ouble val[55][55];struct q{int s,p;};    queue<q> que;double Cal (int p1,int P2) {if (P1 = = 0 && P2 = = 1) return 0;    Pos p1 = Pos[p1];    Pos P2 = pos[p2]; return sqrt ((p1.x+0.0-p2.x) * (p1.x+0.0-p2.x) + (P1.Y+0.0-P2.Y) * (P1.Y+0.0-P2.Y) + 0.0);}    int main () {int n,m;    int i,k,j;    Acautomaton AC; while (scanf ("%d%d", &n,&m) && (n| |        m) {for (i = 1;i <= n; ++i) scanf ("%d%d", &pos[i].x,&pos[i].y); AC.        Init ();            while (m--) {scanf ("%d", &k);            for (i = 0;i < K; ++i) scanf ("%d", &num[i]);            NUM[K] =-1; AC.        Insert (num); } AC.        Getfail (); for (i = 1;i <= N, ++i) {for (j = 1;j <= n; ++j) val[i][j] = sqrt ((pos[i].x-pos[i].x        ) * (pos[i].x-pos[i].x) + (POS[I].Y-POS[I].Y) * (POS[I].Y-POS[I].Y));        } for (i = 1;i <= n; ++i) val[i][0] = Val[0][i] = 0; for (i = 0;i < AC. Top;        ++i) for (j = 1;j <= n; ++j) Dis[i][j] =-1;        Dis[0][0] = 0;        Que.push ((Q) {0,0});        Q f,t; while (que.empty () = = False) {f = Que.front ();           Que.pop ();                for (i = f.p+1;i <= (f.p = = 0? 1:n); ++i) {int tr = F.S;                        while (tr! =-1) {if (st[tr].next[i]! =-1 && st[st[tr].next[i]].flag! = 0)                    Break                tr = St[tr].fail;                } if (tr! =-1) continue;                tr = F.S;                    while (tr! =-1) {if (st[tr].next[i]! =-1) break;                tr = St[tr].fail;                } if (tr = =-1) tr = 0;                else tr = st[tr].next[i]; if (dis[tr][i] = = 1 | | dis[tr][i] > DIS[F.S][F.P] + Cal (f.p,i)) {Dis[tr][i] = dis[f.                    S][F.P] + Cal (f.p,i);                Que.push ((Q) {tr,i});        }}} Double Min =-1; for (i = 0;i < AC.Top;                        ++i) {if (Dis[i][n]! =-1) {if (Min = =-1)                    Min = Dis[i][n];                else min = min (min,dis[i][n]);        }} if (Min = =-1) printf ("Can not is reached!\n");    else printf ("%.2lf\n", Min); } return 0;}

AC automata + two-dimensional shortest path HDU 4511 Xiao Ming series Story--the test of the girlfriend