[JSOI2009]密码

Description

Input

Output

Sample Input
10 2
hello
world

Sample Output
2
helloworld
worldhello

HINT

然后难点就在于如何输出方案

首先42这数字非常妙(生命、宇宙以及任何事情的终极答案)

如果存在一个字符可以任意选的情况，那么答案至少也要为2*26=52，所以这种情况是不存在的

所以就直接爆搜，\(O(n!)\)搜索，然后中间疯狂剪枝就好

/*program from Wolfycz*/
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define inf 0x7f7f7f7f
using namespace std;
typedef long long ll;
typedef unsigned int ui;
typedef unsigned long long ull;
inline char gc(){
    static char buf[1000000],*p1=buf,*p2=buf;
    return p1==p2&&(p2=(p1=buf)+fread(buf,1,1000000,stdin),p1==p2)?EOF:*p1++;
}
inline int frd(){
    int x=0,f=1;char ch=gc();
    for (;ch<'0'||ch>'9';ch=gc())   if (ch=='-')    f=-1;
    for (;ch>='0'&&ch<='9';ch=gc()) x=(x<<1)+(x<<3)+ch-'0';
    return x*f;
}
inline int read(){
    int x=0,f=1;char ch=getchar();
    for (;ch<'0'||ch>'9';ch=getchar())  if (ch=='-')    f=-1;
    for (;ch>='0'&&ch<='9';ch=getchar())    x=(x<<1)+(x<<3)+ch-'0';
    return x*f;
}
inline void print(int x){
    if (x<0)    putchar('-'),x=-x;
    if (x>9)    print(x/10);
    putchar(x%10+'0');
}
const int N=1e2;
struct S1{
    int trie[N+10][26],fail[N+10],End[N+10];
    int root,tot;
    void insert(char *s,int ID){
        int len=strlen(s),p=root;
        for (int i=0;i<len;i++){
            if (!trie[p][s[i]-'a']) trie[p][s[i]-'a']=++tot;
            p=trie[p][s[i]-'a'];
        }
        End[p]=1<<ID;
    }
    void make_fail(){
        static int h[N+10];
        int head=1,tail=0;
        for (int i=0;i<26;i++)  if (trie[root][i])  h[++tail]=trie[root][i];
        for (;head<=tail;head++){
            int Now=h[head];
            End[Now]|=End[fail[Now]];
            for (int i=0;i<26;i++){
                if (trie[Now][i]){
                    int son=trie[Now][i];
                    fail[son]=trie[fail[Now]][i];
                    h[++tail]=son;
                }else   trie[Now][i]=trie[fail[Now]][i];
            }
        }
    }
}AC;//Aho-Corasick automaton
int work(char *s,char *t){
    int lens=strlen(s),lent=strlen(t),Ans=0;
    for (int i=0;i<lens;i++){
        int res=0,x=i,y=0;
        while (x<lens&&y<lent){
            if (s[x]!=t[y]) break;
            res++,x++,y++;
        }
        if (x!=lens)    continue;
        Ans=max(Ans,res);
    }
    return Ans;
}
ll f[30][N+10][(1<<10)+10];
int pos[15],Len[15],g[15][15],L,n,cnt;
bool vis[15];
char s[15][15];
struct S2{char s[N+10];}A[50];
bool operator <(const S2 &x,const S2 &y){
    int lenx=strlen(x.s),leny=strlen(y.s);
    if (lenx!=leny) return lenx<leny;
    for (int i=0;i<lenx;i++)    if (x.s[i]!=y.s[i]) return x.s[i]<y.s[i];
    return 0;
}
void dfs(int x,int len){
    if (len>L)  return;
    if (x==n){
        static char T[N+10];
        for (int i=0;i<Len[pos[0]];i++) T[i]=s[pos[0]][i];
        int L=Len[pos[0]];
        for (int i=1;i<n;i++){
            for (int j=0;j<Len[pos[i]];j++)
                T[j+L-g[pos[i-1]][pos[i]]]=s[pos[i]][j];
            L+=Len[pos[i]]-g[pos[i-1]][pos[i]];
        }
        memcpy(A[cnt++].s,T,sizeof(T));
        return;
    }
    for (int i=0,tmp;i<n;i++){
        if (!vis[i]){
            pos[x]=i,vis[i]=1;
            tmp=len+Len[i]-(x?g[pos[x-1]][i]:0);
            dfs(x+1,tmp);
            pos[x]=0,vis[i]=0;
        }
    }
}
int main(){
    L=read(),n=read();
    for (int i=0;i<n;i++){
        scanf("%s",s[i]);
        Len[i]=strlen(s[i]);
        AC.insert(s[i],i);
    }
    AC.make_fail();
    f[0][0][0]=1;
    for (int i=0;i<L;i++){
        for (int j=0;j<=AC.tot;j++){
            for (int s=0;s<1<<n;s++){
                if (!f[i][j][s])    continue;
                for (int k=0;k<26;k++){
                    int son=AC.trie[j][k];
                    f[i+1][son][s|AC.End[son]]+=f[i][j][s];
                }
            }
        }
    }
    ll Ans=0;
    for (int i=0;i<=AC.tot;i++) Ans+=f[L][i][(1<<n)-1];
    printf("%lld\n",Ans);
    if (Ans>42) return 0;
    for (int i=0;i<n;i++)
        for (int j=0;j<n;j++)
            g[i][j]=work(s[i],s[j]);
    dfs(0,0);
    sort(A,A+cnt);
    for (int i=0;i<cnt;i++) printf("%s\n",A[i].s);
    return 0;
}

原文地址：https://www.cnblogs.com/Wolfycz/p/10486372.html