KMP,AC自动机

KMP

http://www.matrix67.com/blog/archives/115

注意这个从1开始字符串，从0开始不好写，不过不优化fail的从0还是好写- -！

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <algorithm>
 4 #include <iostream>
 5 using namespace std;
 6 #define maxn 1000005
 7 int n, m, k;
 8 int a[maxn],fail[maxn];
 9 char str1[maxn], str2[maxn];
10 void init(){
11     int j = 0;
12     for (int i = 2; i <= m; i++){
13         while (j&&str2[i] != str2[j + 1])j = fail[j];
14         if (str2[i] == str2[j + 1])j++;
15         fail[i] = j;
16     }
17 }
18 void kmp(){
19     int j = 0; k = 0;
20     for (int i = 1; i <= n; i++){
21         while (j&&str1[i] != str2[j + 1])j = fail[j];
22         if (str1[i] == str2[j + 1])j++;
23         if (j == m){ a[k++] = i - m + 1; j = fail[j]; }
24     }
25 }
26 int main(){
27     while (~scanf("%s%s", str1+1, str2+1)){
28         n = strlen(str1 + 1);
29         m = strlen(str2 + 1);
30         init();
31         kmp();
32         for (int i = 0; i < k; i++)printf(i==k-1?"%d\n":"%d ", a[i]);
33     }
34     return 0;
35 }

View Code

AC自动机

http://wenku.baidu.com/link?url=_WLP8BEXP8hSTvQPl_drqiSeOkkEkno0LveEwD5ZRSSb5zolHaV6kVXoELIdrxpWriPjL_O6od4FrMWvpB5dPcX28qbQ0BOd5nBCagt4IN_

首先要把深度为i之前的fail指针求出来才能求深度为i的fail指针，BFS刚好符合！

最后没看懂的再看看下面的这个回答

某某:我把我的想法说下`是不是`对 ``你最后的那个trie先在要求的是she中h的失败指针，而对``h深度`h-1的失败指针都已经求出来（可以通过归纳证明）所以
对h只要沿着它父亲走失败指针的看失败指针指向的节点是否有一个儿子和h 相等，而由失败指针的定义。h以上有n个和失败指针到 root之间
匹配，所以如果找到一个h 说明有n+1个匹配得证。希作者看到后联系我- - 看我的理解对不对``还是有更加具体化的思路我没有想到```

这种代码肯定会被各种卡的，知道写，乱改就可以了。。。我都懒得去知道以前写的了

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <queue>
 4 #include <algorithm>
 5 #include <iostream>
 6 using namespace std;
 7 #define maxn 1005005
 8 #define sz_ 26
 9 int n, k, t, sz, cnt;
10 int m[10005];
11 int q[maxn];
12 int ch[maxn][sz_];
13 int fail[maxn];
14 vector<int>g[10005],val[maxn];
15 char str1[maxn], str2[105];
16 void insert(char *str, int v){
17     int u = 0, len = strlen(str+1);
18     for (int i = 1; i <= len; i++){
19         int c = str[i] - 'a';
20         if (!ch[u][c]){
21             memset(ch[sz], 0, sizeof ch[sz]);
22             val[sz].clear();
23             ch[u][c] = sz++;
24         }
25         u = ch[u][c];
26     }
27     val[u].push_back(v);
28 }
29 void bfs(){
30     int u = 0;
31     int head=0,tail=0;
32     for (int i = 0; i < sz_; i++){
33         int v = ch[u][i];
34         if (v){ fail[v] = 0; q[tail++] = v; }
35     }
36     while (head!=tail){
37         int u = q[head++];
38         for (int i = 0; i < sz_; i++){
39             int v = fail[u];
40             if (!ch[u][i])continue;
41             q[tail++] = ch[u][i];
42             while (v&&!ch[v][i])v = fail[v];
43             fail[ch[u][i]] = ch[v][i];
44         }
45     }
46 }
47 void ac(){
48     int u = 0;
49     for (int i = 1; i <= n; i++){
50         int c = str1[i] - 'a';
51         while (u&&!ch[u][c])u = fail[u];
52         u=ch[u][c];
53         int ux = u;
54         while (ux){
55             for (int j = 0; j < val[ux].size(); j++)
56                 g[val[ux][j]].push_back(i - m[val[ux][j]] + 1);
57             ux = fail[ux];
58         }
59     }
60 }
61 int main(){
62     while (~scanf("%s", str1 + 1)){
63         n = strlen(str1 + 1);
64         scanf("%d", &k);
65         cnt = 0;
66         sz = 1; memset(ch[0], 0, sizeof ch[0]);
67         for (int i = 0; i <= 10004; i++)g[i].clear();
68         for (int i = 1; i <= k; i++){
69             scanf("%s", str2 + 1);
70             m[i] = strlen(str2 + 1);
71             insert(str2, i);
72         }
73         bfs();
74         ac();
75         for (int i = 1; i <= k; i++)cnt+=g[i].size();
76         printf("%d\n", cnt);
77         for (int i = 1; i <= k; i++)
78             for (int j = 0; j < g[i].size(); j++)
79                 printf(j == g[i].size() - 1 ? "%d\n" : "%d ", g[i][j]);
80         for (int i = 0; i <= sz; i++)val[i].clear();
81     }
82     return 0;
83 }

View Code

posted @ 2014-03-31 16:32 HaibaraAi 阅读(133) 评论(0) 收藏举报

刷新页面返回顶部