AC自动机

AC自动机是KMP和Trie的结合，主要处理多模板串匹配问题。下面推荐一个博客，有助于学习AC自动机。

NOTONLYSUCCESS 

这里还有一个Kuangbin开的比赛，大家也可以做一下，加深对算法的理解。

下面是比赛中的题目，采用了notonlysuccess的模板。

HDU 2222 Keywords Search

题意：最裸的模板题，给定一些模板串以及一个文本串，要在文本串中找有多少个模板串。

/*ID: [e-mail]:LANG: C++*/#include<map>#include<set>#include<queue>#include<stack>#include<cmath>#include<cstdio>#include<vector>#include<string>#include<fstream>#include<cstring>#include<ctype.h>#include<iostream>#include<algorithm>#define INF (1<<30)#define PI acos(-1.0)#define mem(a, b) memset(a, b, sizeof(a))#define rep(i, n) for (int i = 0; i < n; i++)#define debug puts("===============")typedef long long ll;using namespace std;const int maxnode = 500100;const int charset = 26;struct ACAutomaton {    int ch[maxnode][charset];    int fail[maxnode];    int Q[maxnode];    int val[maxnode];    int sz;    int ID[128];    //初始化,计算字母对应的儿子ID,如:'a'->0 ... 'z'->25    void init() {        fail[0] = 0;        for (int i = 0; i < charset; i++) ID[i + 'a'] = i;    }    //重新建树需先Reset    void reset() {        memset(ch[0], 0, sizeof(ch[0]));        sz = 1;    }    //将权值为key的字符串a插入到trie中    void Insert(char *s, int key) {        int u = 0;        for ( ; *s; s++) {            int c = ID[*s];            if (!ch[u][c]) {                memset(ch[sz], 0, sizeof(ch[sz]));                val[sz] = 0;                ch[u][c] = sz++;            }            u = ch[u][c];        }        val[u] += key;    }    //建立AC自动机,确定每个节点的权值以及状态转移    void Construct () {        int *s = Q, *e = Q;        for (int i = 0; i < charset; i++) {            if (ch[0][i]) {                fail[ch[0][i]] = 0;                *e ++ = ch[0][i];            }        }        while(s != e) {            int u = *s++;            for (int i = 0; i < charset; i++) {                int &v = ch[u][i];                if (ch[u][i]) {                    *e ++ = v;                    fail[v] = ch[fail[u]][i];                } else {                    v = ch[fail[u]][i];                }            }        }    }    //最基础的查询，询问一个字符串中出现了多少模板串    int query(char *s) {        int ans = 0, u = 0;        for ( ; *s; s++) {            int c = ID[*s];            u = ch[u][c];            int tmp = u;            while(tmp) {                ans += val[tmp];                val[tmp] = 0;                tmp = fail[tmp];            }        }        return ans;    }}AC;char str[1000100];int main() {    AC.init();    int t, n;    scanf("%d", &t);    while(t--) {        scanf("%d", &n);        AC.reset();        for (int i = 0; i < n; i++) {            scanf("%s", str);            AC.Insert(str, 1);        }        AC.Construct();        scanf("%s", str);        printf("%d\n", AC.query(str));    }    return 0;}

HDU 2896 病毒侵袭

题意：有N个病毒，M个文本串，问每个文本串出现了多少个病毒，分别是哪些？一共有多少个文本串出现了病毒？

思路：这道题的病毒可以包含所有可见ASC码值

/*ID: [e-mail]:LANG: C++*/#include<map>#include<set>#include<queue>#include<stack>#include<cmath>#include<cstdio>#include<vector>#include<string>#include<fstream>#include<cstring>#include<ctype.h>#include<iostream>#include<algorithm>#define INF (1<<30)#define PI acos(-1.0)#define mem(a, b) memset(a, b, sizeof(a))#define rep(i, n) for (int i = 0; i < n; i++)#define debug puts("===============")typedef long long ll;using namespace std;const int maxnode = 100100;const int charset = 128;struct ACAutomaton {    int ch[maxnode][charset];    int fail[maxnode];    int Q[maxnode];    int val[maxnode];    int sz;    int ID[128];    void init() {        fail[0] = 0;        for (int i = 0; i < charset; i++) ID[i] = i;    }    void reset() {        sz = 1;        memset(ch[0], 0, sizeof(ch[0]));    }    void Insert(char *s, int key) {        int u = 0;        for ( ; *s; s++) {            int c = ID[*s];            if (!ch[u][c]) {                memset(ch[sz], 0, sizeof(ch[sz]));                val[sz] = 0;                ch[u][c] = sz++;            }            u = ch[u][c];        }        val[u] = key;    }    void Construct () {        int *s = Q, *e = Q;        for (int i = 0; i < charset; i++) {            if (ch[0][i]) {                *e++ = ch[0][i];                fail[ch[0][i]] = 0;            }        }        while(s != e) {            int u = *s++;            for (int i = 0; i < charset; i++) {                int &v = ch[u][i];                if (v) {                    *e++ = v;                    fail[v] = ch[fail[u]][i];                } else {                    v = ch[fail[u]][i];                }            }        }    }    void query(char *s, int &tot, int id) {        int ans = 0, u = 0;        set<int> S;        set<int>::iterator it;        S.clear();        for (; *s; s++) {            int c = ID[*s];            u = ch[u][c];            int tmp = u;            while(tmp) {                if (val[tmp]) S.insert(val[tmp]), ans++;                tmp = fail[tmp];            }        }        if (ans) {            printf("web %d:", id);            for (it = S.begin(); it != S.end(); it++) printf(" %d", *it);            putchar('\n');            tot++;        }    }}AC;char buf[210], str[10100];int main () {    int n, m, tot = 0;    scanf("%d", &n);    AC.init();    AC.reset();    for (int i = 0; i < n; i++) {        scanf("%s", buf);        AC.Insert(buf, i + 1);    }    AC.Construct();    scanf("%d", &m);    for (int i = 0; i < m; i++) {        scanf("%s", str);        AC.query(str, tot, i + 1);    }    printf("total: %d\n", tot);    return 0;}

HDU 3065 病毒侵袭持续中

题意：有N个病毒，一个文本串，问文本串中每个病毒出现了多少次

思路：也是基础的模板，是多case。。

/*ID: [e-mail]:LANG: C++*/#include<map>#include<set>#include<queue>#include<stack>#include<cmath>#include<cstdio>#include<vector>#include<string>#include<fstream>#include<cstring>#include<ctype.h>#include<iostream>#include<algorithm>#define INF (1<<30)#define PI acos(-1.0)#define mem(a, b) memset(a, b, sizeof(a))#define rep(i, n) for (int i = 0; i < n; i++)#define debug puts("===============")typedef long long ll;using namespace std;const int maxnode = 50010;const int charset = 128;int cnt[1100];struct ACAutomaton {    int ch[maxnode][charset];    int fail[maxnode];    int Q[maxnode];    int val[maxnode];    int sz;    int ID[128];    void init() {        fail[0] = 0;        for (int i = 0; i < charset; i++) ID[i] = i;    }    void reset() {        sz = 1;        memset(ch[0], 0, sizeof(ch[0]));    }    void Insert(char *s, int key) {        int u = 0;        for ( ; *s; s++) {            int c = ID[*s];            if (!ch[u][c]) {                memset(ch[sz], 0, sizeof(ch[sz]));                val[sz] = 0;                ch[u][c] = sz++;            }            u = ch[u][c];        }        val[u] = key;    }    void Construct () {        int *s = Q, *e = Q;        for (int i = 0; i < charset; i++) {            if (ch[0][i]) {                *e++ = ch[0][i];                fail[ch[0][i]] = 0;            }        }        while(s != e) {            int u = *s++;            for (int i = 0; i < charset; i++) {                int &v = ch[u][i];                if (v) {                    *e++ = v;                    fail[v] = ch[fail[u]][i];                } else {                    v = ch[fail[u]][i];                }            }        }    }    void query(char *s) {        int u = 0;        for (; *s; s++) {            int c = ID[*s];            u = ch[u][c];            int tmp = u;            while(tmp) {                if (val[tmp]) cnt[val[tmp]]++;                tmp = fail[tmp];            }        }    }} AC;char buf[1100][55], str[2000100];int main () {    int n, m, tot = 0;    AC.init();    while(~scanf("%d", &n)) {        AC.reset();        for (int i = 0; i < n; i++) {            scanf("%s", buf[i]);            AC.Insert(buf[i], i + 1);            cnt[i + 1] = 0;        }        AC.Construct();        scanf("%s", str);        AC.query(str);        for (int i = 1; i <= n; i++) if (cnt[i]) printf("%s: %d\n", buf[i - 1], cnt[i]);    }    return 0;}

ZOJ 3430 Detect the Virus

题意：有一种编码方式，将输进来的字符转化为二进制，然后6个为一组，不足补零，得到一个新的数字，每个数字对应一个字符（见题面）。现在给你已经编码过的n个病毒，和m个编码过的文本串，问每个文本串各包含多少种病毒。

思路：这里反编码的时候，会发现可能有256种状态，所以不能用字符串表示。反编码之后就是裸的AC自动机。

/*ID: [e-mail]:LANG: C++*/#include<map>#include<set>#include<queue>#include<stack>#include<cmath>#include<cstdio>#include<vector>#include<string>#include<fstream>#include<cstring>#include<ctype.h>#include<iostream>#include<algorithm>#define INF (1<<30)#define PI acos(-1.0)#define mem(a, b) memset(a, b, sizeof(a))#define rep(i, n) for (int i = 0; i < n; i++)#define debug puts("===============")typedef long long ll;using namespace std;const int maxnode = 510 * 64;const int charset = 256;struct ACAutomaton {    int ch[maxnode][charset];    int fail[maxnode];    int Q[maxnode];    int val[maxnode];    int sz;    int ID[256];    void init() {        fail[0] = 0;        //for (int i = 0; i < charset; i++) ID[i] = i;    }    void reset() {        sz = 1;        memset(ch[0], 0, sizeof(ch[0]));    }    void Insert(unsigned char s[], int key, int len) {        int u = 0;        for (int i = 0; i < len; i++) {            int c = s[i];            if (!ch[u][c]) {                memset(ch[sz], 0, sizeof(ch[sz]));                val[sz] = 0;                ch[u][c] = sz++;            }            u = ch[u][c];        }        val[u] = key;    }    void Construct () {        int *s = Q, *e = Q;        for (int i = 0; i < charset; i++) {            if (ch[0][i]) {                *e++ = ch[0][i];                fail[ch[0][i]] = 0;            }        }        while(s != e) {            int u = *s++;            for (int i = 0; i < charset; i++) {                int &v = ch[u][i];                if (v) {                    *e++ = v;                    fail[v] = ch[fail[u]][i];                } else {                    v = ch[fail[u]][i];                }            }        }    }    void query(unsigned char s[], int len) {        int u = 0, ans = 0;        bool vis[520] = {0};        for (int i = 0; i < len; i++) {            int c = s[i];            u = ch[u][c];            int tmp = u;            while(tmp) {                if (val[tmp] && !vis[val[tmp]]) {                    ans++, vis[val[tmp]] = 1;                }                tmp = fail[tmp];            }        }        printf("%d\n", ans);    }} AC;char s[4000];unsigned char g[4000];unsigned char now[4000];void get(char *s, int len) {    for (int i = 0; i < len; i++) {        if (s[i] >= 'A' && s[i] <= 'Z') g[i] = s[i] - 'A';        else if (s[i] >= 'a' && s[i] <= 'z') g[i] = s[i] - 'a' + 26;        else if (s[i] >= '0' && s[i] <= '9') g[i] = s[i] - '0' + 52;        else if (s[i] == '+') g[i] = 62;        else g[i] = 63;    }    g[len] = 0;}int change(unsigned char g[], int len) {    int cnt = 0;    for (int i = 0; i < len; i += 4) {        now[cnt++] = (g[i] << 2) | (g[i + 1] >> 4);        if (i + 2 < len) now[cnt++] = (g[i + 1] << 4) | (g[i + 2] >> 2);        if (i + 3 < len) now[cnt++] = (g[i + 2] << 6) | g[i + 3];    }    return cnt;}int main () {    int n, m;    AC.init();    while(~scanf("%d", &n)) {        AC.reset();        for (int i = 0; i < n; i++) {            scanf("%s", s);            int len = strlen(s);            while(s[len - 1] == '=') len--;            get(s, len);            int cnt = change(g, len);            AC.Insert(now, i + 1, cnt);        }        AC.Construct();        scanf("%d", &m);        while(m--) {            scanf("%s", s);            int len = strlen(s);            while(s[len - 1] == '=') len--;            get(s, len);            int cnt = change(g, len);            AC.query(now, cnt);        }        putchar('\n');    }    return 0;}

POJ 2778 DNA Sequence

题意：DNA的序列由ACTG四个字母组成，现在给定m个不可行的序列。问随机构成的长度为n的序列中，有多少种序列是可行的（只要包含一个不可行序列便不可行）。个数很大，对100000取模。 

思路：AC自动机 + DP 解题报告

HDU 2243 考研路茫茫――单词情结

题意：给定一些词根，如果一个单词包含有词根，则认为是有效的。现在问长度不超过L的单词里面，有多少有效的单词？

解题报告

ZOJ 2619 Generator

题意：给定一个数N，代表可以选前N个字母。然后给定一个仅有前N个字母组成的字符串，问从空串开始构造，每次可以在已有基础上从前N个字母中挑选一个加在后面，问构造的字符串的长度期望是多少？

解题报告

持续更新中