2782: [HNOI2006] Shortest female string

2782: [HNOI2006] Shortest mother string Time limit:1 Sec Memory limit:128 MB
Submit:3 Solved:2
[Submit] [Status] [Web Board] Description given n strings (S1,s2, ", SN), requires finding a shortest string t, so that n strings (S1,s2,", SN) are substrings of T. The first line of input is a positive integer n (n<=12) that represents the number of strings given. The following n lines, each with a string of all uppercase letters. The length of each string does not exceed 50. Output has only one row, which is the shortest string found for T. In the shortest possible case, if more than one string satisfies the requirement, the first in the dictionary order must be output. Sample Input

2ABCDBCDABC

Sample Output

Abcdabc

HINT Source

Exercises

　　

First of all, my thinking is very stupid, isf[i][J][k] F[I][J][K] indicates a length ofII, the node on the automaton is JJ, it is impossible to include a substring with a state of KK. But this complexity is directly exploding.

But we can makef [i ][j ]F[i][j] indicates that the node on the automaton is iI, which contains the Shortest length when the status of the substring is J J.

So we turned into the Benquan are 1 of the single source shortest, BFS on the line.

If the BFS is carried out in alphabetical order from small to large, the answer to the smallest dictionary order is obtained.

(a bit of card memory)

#include <iostream>#include<cstdio>#include<cstring>#include<queue>#include<algorithm>using namespacestd;Const intinf=1e9;structnode{intfail,ch[ -],val; voidClear () {fail=val=0; memset (CH,0,sizeof(CH)); }}tr[605];intsz;intbin[ -],n;voidInsertChar*s,intval) {    intu=0, len=strlen (s);  for(intI=0; i<len;i++){        intc=s[i]-'A'; if(!Tr[u].ch[c]) {Tr[u].ch[c]=++sz;        Tr[sz].clear (); } u=Tr[u].ch[c]; } tr[u].val|=bin[val];}voidbuild () {Queue<int>q;q.push (0);  while(!Q.empty ()) {        intu=Q.front (); Q.pop ();  for(intI=0;i< -; i++){            if(Tr[u].ch[i]) {intv=Tr[u].ch[i]; if(u) tr[v].fail=Tr[tr[u].fail].ch[i];            Q.push (v); }Elsetr[u].ch[i]=Tr[tr[u].fail].ch[i]; }    }}intfromu[605][4100],froms[605][4100];BOOLvis[605][4100];voidPrintintUints) {    if(!u)return;    Print (Fromu[u][s],froms[u][s]);  for(intI=0;i< -; i++){        intv=tr[fromu[u][s]].ch[i],t=Froms[u][s];  for(intp=v;p;p=tr[p].fail) T|=Tr[p].val; if(v==u&&t==s) {Putchar ('A'+i);  Break; }    }}voidBFs () {Queue<int>q1,q2; Q1.push (0); Q2.push (0); vis[0][0]=1;  while(!Q1.empty ()) {        intU=q1.front (), s=Q2.front (); Q1.pop (); Q2.pop ();  for(intI=0;i< -; i++){            intv=tr[u].ch[i],t=R;  for(intp=v;p;p=tr[p].fail) T|=Tr[p].val; if(!Vis[v][t]) {Vis[v][t]=1;                Q1.push (v); Q2.push (t); Fromu[v][t]=u; Froms[v][t]=s; if(t==bin[n+1]-1) {print (v,t); return; }            }        }    }}Chars[ -];intMain () {scanf ("%d",&N); Tr[sz=0].clear (); bin[1]=1;  for(intI=2;i< -; i++) bin[i]=bin[i-1]<<1;  for(intI=1; i<=n;i++) {scanf ("%s", s);    Insert (s,i);    } build ();    BFS (); return 0;}

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2782: [HNOI2006] Shortest female string