Aho_Corasick 的一些练习(hdu 2222; Poj 1625; Poj2778; hdu2457; Hdu3247)

其实也没有做很多题，但还是发出来（能骗骗访问量。。。）
Hdu 2222

//本代码有误（有反例，　但是ａｃ了）……但希望有人能告诉我真的是数据水吗。。。。
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <queue>

const int SZ = 26;
const int MX_N = 1e6 + 10;
#define FILL(a, b) memset(a, b, sizeof a)
#define rep(i, s, t) for(register int i = s; i <= t; ++i)

using namespace std;

template <class T>
T read(T x = 0) { 
    char c = getchar();
    while(c < '0' || c > '9') c = getchar();
    while(c >= '0' && c <= '9') 
        x = x*10 + c-'0', c = getchar();
    return x;
}

namespace Aho_Corasick {
    queue<int> Q;
    int ch[MX_N][SZ], tot;
    int val[MX_N], fail[MX_N];

    void init() {
        FILL(fail, 0), tot = 0;
        FILL(val, 0), FILL(ch, 0);
    }

    void insert(char *s) {
        int u = 0;
        for(; *s; ++s) {
            int x = *s - 'a';
            if(!ch[u][x]) ch[u][x] = ++tot;
            u = ch[u][x];
        }++val[u];
    }

    void Bfs_fail() {
        Q.push(0);
        while(!Q.empty()) {
            int u = Q.front(), v; Q.pop();
            rep(i, 0, 25) if((v=ch[u][i])) {
                int t = fail[u];
                if(u) {
                    for(; t && !ch[t][i]; t = fail[t]);
                    fail[v] = ch[t][i];
                }Q.push(v);
            }else ch[u][i] = ch[fail[u]][i];
        }
    }

    int Query(char *T) {
        int u = 0, ret = 0;
        for(; *T; ++T) {
            int x = *T - 'a';
            for(int v = (u=ch[u][x]); val[v]; v = fail[v]) 
                ret += val[v], val[v] = 0;
                //这个地方错了，　这个代码是错的，ｆｏｒ循环的终止条件改成“ｖ”，这样思路上没错但好像Ｔ了，　应该要对访问过的点记一个ｍａｒｋ才算真正过了；
        }return ret;
    }
}using namespace Aho_Corasick;

int main() {
#ifndef ONLINE_JUDGE
    freopen("input.in", "r", stdin);
    freopen("res.out", "w", stdout);
#endif

    int T = read<int>();
    char s[55], t[MX_N];
    while(T--) {
        init();
        int n = read<int>();
        rep(i, 1, n) scanf("%s", s), insert(s);
        Bfs_fail();
        scanf("%s", t);
        printf("%d\n", Query(t));
    }
    return 0;
}
//模板题代码就是给大家参考一下

POJ 1625 Censored!

/*
给n个字母，构成长度为m的串，总共有n^m种。
给p个字符串，问n^m种字符串中不包含（不是子串）这p个字符串的个数。
*/
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <queue>

const int SZ = 1e2 + 10;
const int MX_N = 50 + 10;
const int MX_Node = 3e2 + 10;
#define US unsigned char
#define Fill(a, b) memset(a, b, sizeof a)
#define rep(i, s, t) for(register int i = s; i <= t; ++i)
#define dec(i, s, t) for(register int i = s; i >= t; --i)

using namespace std;

template <class T>
T read(T x = 0) {
    char c = getchar();
    while(c < '0' || c > '9') c = getchar();
    while(c >= '0' && c <= '9') 
        x = x*10 + c-'0', c = getchar();
    return x;
}

int n, m, q;
namespace Aho_Corasick {
    int tot;
    queue<int> Q;
    int ch[MX_Node][SZ], val[MX_Node], fail[MX_Node];

    int Map, hash[MX_Node];
    void Hash(US *s) {
        Map = n-1;
        rep(i, 0, Map) hash[s[i]] = i;
    }

    void init() {
        tot = 0;
        Fill(ch, 0), Fill(val, 0);
        Fill(fail, 0), Fill(hash, 0);
    }

    void insert(US *s) {
        int u = 0;
        for(; *s; ++s) {
            int x = hash[*s];
            if(!ch[u][x]) ch[u][x] = ++tot;
            u = ch[u][x];
        }val[u] = 1;
    }

    void Bfs_fail() {
        Q.push(0);
        while(!Q.empty()) {
            int u = Q.front(), v; Q.pop();
            rep(i, 0, Map) if((v=ch[u][i])) { 
                int t = fail[u];
                if(u) {
                    for(; t && !ch[t][i]; t = fail[t]);
                    fail[v] = ch[t][i];
                }Q.push(v);
            }else ch[u][i] = ch[fail[u]][i];
            val[u] |= val[fail[u]];
        }
    }
}using namespace Aho_Corasick;

struct Bn {
    int x[MX_N], len;
    Bn() { memset(x, 0, sizeof x), len = 0; }

    void print() {
        printf("%d", x[len]);
        dec(i, len-1, 1) printf("%.4d", x[i]);
        puts("");
    }
}temp;

const int Mod = 1e4;
Bn operator + (Bn a, Bn b) {
    Fill(temp.x, 0);
    temp.len = max(a.len, b.len);
    rep(i, 1, temp.len) {
        temp.x[i] += a.x[i] + b.x[i];
        if(temp.x[i] >= Mod) 
            temp.x[i+1] += temp.x[i] / Mod, temp.x[i] %= Mod;
    }while(temp.x[temp.len+1]) ++temp.len;
    return temp;
}

Bn f[MX_N][MX_Node], Ans;
void solve() {
    rep(i, 0, MX_N-1)rep(j, 0, MX_Node-1) 
        Fill(f[i][j].x, 0), f[i][j].len = 1;
    Fill(Ans.x, 0), Ans.len = 1;

    f[0][0].x[1] = 1, f[0][0].len = 1;

    rep(i, 1, m)rep(u, 0, tot) if(!val[u])
        rep(k, 0, Map) {
            int v = ch[u][k];
            if(!val[v]) f[i][v] = f[i][v] + f[i-1][u];
        }
    rep(i, 0, tot) if(!val[i])
        Ans = Ans + f[m][i];

    Ans.print();
}

int main() {
#ifndef ONLINE_JUDGE
    freopen("input.in", "r", stdin);
    freopen("res.out", "w", stdout);
#endif

    US s[MX_N];
    while(scanf("%d%d%d", &n, &m, &q) == 3) {
        init();
        scanf("%s", s), Hash(s);

        while(q--) {
            scanf("%s", s);
            insert(s);
        }Bfs_fail();

        solve();
    }
    return 0;
}
//比较恶心的是要写高精，以及字符串的读入必须用ｕｎｓｉｇｎｅｄ char不然会弄出负数
//状态的设置f[i][j]　->　len = i, 以自动机的j节点结尾的方案
//MX_N, MX_Node这些写得好像有点丑不要介意

POJ 2778. DNA Sequence

//同上题，只是输入格式及数据范围不同，以及给了模数！！！！
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <queue>

const int Mod = 1e5;
const int Sz = 3;
const int MX_Node = 1e2 + 10;
#define Fill(a, b) memset(a, b, sizeof a)
#define rep(i, s, t) for(register int i = s; i <= t; ++i)
#define dec(i, s, t) for(register int i = s; i >= t; --i)

using namespace std;

template <class T>
T read(T x = 0) {
    char c = getchar();
    while(c < '0' || c > '9') c = getchar();
    while(c >= '0' && c <= '9') 
        x = x*10 + c-'0', c = getchar();
    return x;
}

namespace Aho_Corasick {
    int tot;
    queue<int> Q;
    int ch[MX_Node][Sz + 1], val[MX_Node], fail[MX_Node];

    int hash[MX_Node<<1];
    void Hash() {
        hash['T'] = 0, hash['A'] = 1;
        hash['G'] = 2, hash['C'] = 3;
    }

    void init() {
        tot = 0;
        Fill(ch, 0), Fill(val, 0);
        Fill(fail, 0), Fill(hash, 0);
    }

    void insert(char *s) {
        int u = 0;
        for(; *s; ++s) {
            int x = hash[*s];
            if(!ch[u][x]) ch[u][x] = ++tot;
            u = ch[u][x];
        }val[u] = 1;
    }

    void Bfs_fail() {
        Q.push(0);
        while(!Q.empty()) {
            int u = Q.front(), v; Q.pop();
            rep(i, 0, Sz) if((v=ch[u][i])) { 
                if(u) {
                    int t = fail[u];
                    for(; t && !ch[t][i]; t = fail[t]);
                    fail[v] = ch[t][i];
                }Q.push(v);
            }else ch[u][i] = ch[fail[u]][i];
            val[u] |= val[fail[u]];
        }
    }

}using namespace Aho_Corasick;

struct Matrix {
    long long G[110][110];
    Matrix() {Fill(G, 0);}
}unit;

Matrix operator * (Matrix a, Matrix b) {
    Matrix c;
    rep(i, 0, tot)rep(j, 0, tot) {
        rep(k, 0, tot) 
            c.G[i][j] += a.G[i][k] * b.G[k][j];
        c.G[i][j] %= Mod;
    }
    return c;
}
Matrix operator ^ (Matrix a, long long b) {
    Matrix ret = a;
    for(--b; b; b >>= 1LL, a = a*a)
        if(b & 1LL) ret = ret * a;
    return ret;
}

void build_matrix() {
    rep(i, 0, tot)rep(j, 0, Sz)
        unit.G[i][ch[i][j]] += !val[i] && !val[ch[i][j]];
}

int n;
long long m;
int main() {
#ifndef ONLINE_JUDGE
    freopen("input.in", "r", stdin);
    freopen("res.out", "w", stdout);
#endif

    static char s[MX_Node];
    scanf("%d%lld", &n, &m);
    Hash();
    while(n--) {
        scanf("%s", s);
        insert(s);
    }
    Bfs_fail();

    build_matrix();
    unit = unit ^ m;

    long long Ans = 0;
    rep(i, 0, tot) Ans = (Ans + unit.G[0][i]) % Mod;
    printf("%lld\n", Ans);
    return 0;
}
//“类似”flyod矩阵的一种东西，　方案数可以用矩阵累加，　其实就是对于矩阵里的每个点，　就是计算从别的点累计过来的方案

HDU 2457. DNA repair/POJ 3691

//给出一些DNA病毒串及一个基因序列，求最少的修改使基因序列不包含病毒串
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <queue>

const int oo = 1e8;
const int size = 4;
const int mx_n = 1e3 + 10;
const int mx_node = 5e2 + 10;
#define Fill(a, b) memset(a, b, sizeof a)
#define rep(i, s, t) for(register int i = s; i <= t; ++i)

using namespace std;

namespace Aho_Corasick {
    queue<int> Q;
    int tot, h[mx_node];
    int ch[mx_node][size], val[mx_node], fail[mx_node];

    void init() {
        tot = 0, Fill(ch, 0);
        Fill(val, 0), Fill(fail, 0);
    }

    void Hash() {
        h['C'] = 0;
        h['A'] = 1;
        h['T'] = 2;
        h['G'] = 3;
    }

    void insert(char *s) {
        int u = 0;
        for(; *s; ++s) {
            int x = h[*s];
            if(!ch[u][x]) ch[u][x] = ++tot;
            u = ch[u][x];
        }val[u] = 1;
    }

    void Bfs_fail() {
        Q.push(0);
        while(!Q.empty()) {
            int u = Q.front(), v; Q.pop();
            rep(i, 0, size-1) if((v=ch[u][i])) {
                if(u) {
                    int t = fail[u];
                    for(; t && !ch[t][i]; t = fail[t]);
                    fail[v] = ch[t][i];
                }Q.push(v);
            }else ch[u][i] = ch[fail[u]][i];
            val[u] |= val[fail[u]];
        }
    }
}using namespace Aho_Corasick;

char T[mx_n];
int f[mx_n][mx_node];
void solve() {
    scanf("%s", T);
    int len = strlen(T), Ans = oo;
    rep(i, 0, len)rep(j, 0, tot) 
        f[i][j] = oo; f[0][0] = 0;

    rep(i, 0, len-1) {
        int idx = h[T[i]], v;
        rep(u, 0, tot) if(!val[u] && f[i][u] ^ oo)
            rep(x, 0, size-1) if(!val[(v=ch[u][x])]) {
                if(x ^ idx) f[i+1][v] = min(f[i][u] + 1, f[i+1][v]);
                else f[i+1][v] = min(f[i][u], f[i+1][v]);
            }
    }rep(i, 0, tot) Ans = min(Ans, f[len][i]);
    printf("%d\n", Ans <= len? Ans : -1);
}

int main() {
#ifndef ONLINE_JUDGE
    freopen("input.in", "r", stdin);
    freopen("res.out", "w", stdout);
#endif
    int Case = 0, n;
    Hash();
    while(scanf("%d", &n) && n) {
        char s[50];
        init();

        while(n--) {
            scanf("%s", s);
            insert(s);
        }Bfs_fail();

        printf("Case %d: ", ++Case);
        solve();
    }
    return 0;
}
//对于状态的设置　f[i][j], 
//len = i　的串,　在ａｃ自动机上能匹配到以节点ｊ结尾的最少修改数
//转移的话比较显然。。。。。

Hdu3247

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <queue>

const int oo = 5e8;
const int mx_n = 12;
const int mx_s = (1 << 11) + 1;
const int mx_node = 6e4 + 10;
#define Fill(a, b) memset(a, b, sizeof a)
#define rep(i, s, t) for(register int i = s; i <= t; ++i)

using namespace std;

int n, m;
namespace Aho_Corasick {
    queue<int> Q;
    int tot, fail[mx_node];
    int ch[mx_node][2], val[mx_node][2];

    void init() {
        tot = 0;
        Fill(fail, 0), Fill(ch, 0), Fill(val, 0);
    }

    void insert(char *s, int k) {
        int u = 0;
        for(; *s; ++s) {
            int x = *s - '0';
            if(!ch[u][x]) ch[u][x] = ++tot;
            u = ch[u][x];
        }if(!k) val[u][0] = 1;
        else val[u][1] = 1 << k;
    }

    void Bfs_fail() {
        Q.push(0);
        while(!Q.empty()) {
            int u = Q.front(), v; Q.pop();
            rep(i, 0, 1) if((v=ch[u][i])) {
                if(u) fail[v] = ch[fail[u]][i];
                Q.push(v);
            }else ch[u][i] = ch[fail[u]][i];
            val[u][0] |= val[fail[u]][0];
            val[u][1] |= val[fail[u]][1];
        }
    }
}using namespace Aho_Corasick;

int G[mx_n][mx_n], dis[mx_node];
int f[mx_s][mx_n], p[mx_n], sz;
void Pre_dist(int u) {
    rep(i, 0, tot) dis[i] = oo;
    Q.push(p[u]), dis[p[u]] = 0;

    while(!Q.empty()) {
        int u = Q.front(), v; Q.pop();
        rep(i, 0, 1) 
            if(dis[v=ch[u][i]] >= oo && !val[v][0])
                dis[v] = dis[u] + 1, Q.push(v);
    }rep(i, 0, sz) G[u][i] = dis[p[i]];
    return ;
}

void solve() {
    rep(i, 0, (1<<(n+1))-1)rep(j, 0, sz) 
        f[i][j] = oo; 
    f[1][0] = 0;

    rep(i, 1, (1<<(n+1))-1)
        rep(j, 0, sz) {
            if(f[i][j] >= oo) continue;
            rep(k, 0, sz)
                f[i|val[p[k]][1]][k] = min(f[i|val[p[k]][1]][k], f[i][j] + G[j][k]);
        }
    int Ans = oo;
    rep(i, 0, sz) Ans = min(Ans, f[(1<<(n+1))-1][i]);
    printf("%d\n", Ans);
    return ;
}

int main() {
#ifndef ONLINE_JUDGE
    freopen("input.in", "r", stdin);
    freopen("res.out", "w", stdout);
#endif

    char s[1000 + 10];
    while(scanf("%d%d", &n, &m) == 2 && n, m) {
        init();

        rep(i, 1, n) scanf("%s", s), insert(s, i);
        rep(i, 1, m) scanf("%s", s), insert(s, 0);
        Bfs_fail();

        sz = 0, Fill(G, 0);
        rep(i, 1, tot) if(val[i][1]) 
            p[++sz] = i;
        rep(i, 0, sz) Pre_dist(i);

        solve();
    }
    return 0;
}
//状压转移，用ＢＦＳ求出两串不过病毒的最短距离，记得要加入Ｒｏｏｔ不然会出问题的
//提供一份思路比较清晰的代码，虽然我的状态是从１开始存的