UVALive 4126 Password Suspects (AC自动机+DP)

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题目链接：https://vjudge.net/problem/UVALive-4126

题目大意：有一个长度为n的未知小写字母串，你已经知道了它的一些连续子串（但不知道出现位置，这些字串可能相互重叠）。比如，若长度为10，有两个连续子串hello和world，则只有两种可能：helloworld和worldhello。求可能的串的个数，若个数不超过42，按字典序输出所有串。

思路：将连续子串构造AC自动机。然后在自动机上进行DP，令dp(u, step, S)表示当前在结点u，已经走了step步（即已经造出了长度为step的串），包含的连续子串集合为S，接着往下构造的方案数，答案即为dp(0, 0, 0)。输出方案时同样按照DP方程的思路进行转移即可。

#include<cstdio>#include<cstring>#include<string>#include<cctype>#include<iostream>#include<set>#include<map>#include<cmath>#include<sstream>#include<vector>#include<stack>#include<queue>#include<algorithm>#define fin freopen("a.txt","r",stdin)#define fout freopen("a.txt","w",stdout)typedef long long LL;typedef unsigned long long ULL;using namespace std;const int inf = 1e9 + 10;const int maxnode = 105;const int sigma_size = 26;const int maxn = 5e4 + 10;int Max;char s[100];LL dp[maxnode][maxnode][1<<10];int n, m;struct Tree{int ch[maxnode][sigma_size];int f[maxnode];int val[maxnode];int last[maxnode];int sz;int idx(char c) { return c - 'a'; }    void init() {        memset(ch[0], 0, sizeof ch[0]);        sz = 1;    }    void getFail() {        queue<int> q;        f[0] = 0;        for(int c = 0; c < sigma_size; c++) {        int u = ch[0][c];        if(u) { f[u] = 0; q.push(u); last[u] = 0;}        }        while(!q.empty()) {        int r = q.front(); q.pop();        for(int c = 0; c < sigma_size; c++) {        int u = ch[r][c];        if(!u) { ch[r][c] = ch[f[r]][c]; continue; }        q.push(u);        int v = f[r];        while(v && !ch[v][c]) v = f[v];        f[u] = ch[v][c];        last[u] = val[f[u]] ? f[u] : last[f[u]];        }        }    }    bool insert(char *s, int v) {    bool ok = false;    int n = strlen(s), u = 0;    for(int i = 0; i < n; i++) {    int c = idx(s[i]);    if(!ch[u][c]) {    memset(ch[sz], 0, sizeof ch[sz]);    val[sz] = 0;    ch[u][c] = sz++;    ok = true;    }    u = ch[u][c];    }    if(ok) val[u] = v;    return ok;    }    void print(int u, int& S)    {        if(u) {            S |= (1<<(val[u]-1));            print(last[u], S);        }    }    LL DP(int u, int step, int S)    {        if(step == n) { return S == (1<<m)-1; }        LL &ans = dp[u][step][S];        if(ans >= 0) return ans;        ans = 0;        for(int i = 0; i < sigma_size; i++)        {            int v = ch[u][i];            int S2 = S;            if(val[v]) print(v, S2);            else if(last[v]) print(last[v], S2);            ans += DP(v, step+1, S2);        }        return ans;    }    void Out(int u, int step, int S, string s)    {    if(step == n) { printf("%s\n", s.c_str()); return ;}        for(int i = 0; i < sigma_size; i++)        {        int v = ch[u][i];        int S2 = S;        if(val[v]) print(v, S2);        else if(last[v]) print(last[v], S2);        if(DP(v, step+1, S2)) Out(v, step+1, S2, s + char('a'+i));        }    }}ac;int main() {   int kase = 0;   while(scanf("%d%d", &n, &m) && n) {      ac.init();      memset(dp, -1, sizeof dp);      for(int i = 1; i <= m; i++) {          scanf("%s", s);          if(!ac.insert(s, i)) { m--; i--; }      }      ac.getFail();      LL ans = ac.DP(0, 0, 0);      printf("Case %d: %lld suspects\n", ++kase, ans);      if(ans <= 42) { ac.Out(0, 0, 0, ""); }   }   return 0;}

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