SPOJ NSUBSTR

考察对father链的理解……

求长度为k的子串，出现次数最多的，出现多少次。

然后强行询问len次， 分别为1..len。

1、建立后缀自动机。

2、显然我们有若干个状态（我对状态的理解不够深刻），

对于主状态（就是整个串的最长链）上任意一个状态，如果到了这个状态意味着什么？

3、意味着，这个状态的所有father实际上也到达了。 

4、根据后缀自动机的性质， 一个状态跳回的father状态，那么一定可以到father状态的len。 （我都在说一些什么啊！）

5、拓扑排序，倒着DP即可。

6、一旦我们能得到一个长度为8的合法串出现次数，那么长度为7的合法串出现的次数…… 可以是长为8的串里长为7的部分！ 

我都看不懂我再说什么！

#include <iostream>#include <cstring>#include <cstdlib>#include <cstdio>using namespace std;const int CHAR = 26;const int MAXN = 250000 * 2  + 100;char str[MAXN];bool duanyan(bool flag){if (!flag){throw 0;}}struct SAM_Node{SAM_Node *fa, *next[CHAR];int len;int id, pos;int dp;SAM_Node(){}SAM_Node(int _len){fa = 0;len = _len;dp=0;memset(next, 0, sizeof(next));}};SAM_Node SAM_node[MAXN *2], *SAM_root, *SAM_last;int SAM_size;SAM_Node * newSAM_Node(int len){SAM_node[SAM_size] = SAM_Node(len);SAM_node[SAM_size].id = SAM_size; SAM_node[SAM_size].dp = 0; return &SAM_node[SAM_size++];}SAM_Node *newSAM_Node(SAM_Node *p){SAM_node[SAM_size] = *p;SAM_node[SAM_size].id = SAM_size;SAM_node[SAM_size].dp = 0; return &SAM_node[SAM_size++];}void SAM_init(){SAM_size = 0;SAM_root = SAM_last = newSAM_Node(0);SAM_node[0].pos = 0;}void SAM_add(int x, int len){SAM_Node *p = SAM_last, *np = newSAM_Node(p -> len+1);np -> pos = len;SAM_last = np;for (;p && !p -> next[x]; p = p -> fa)p -> next[x] = np;if (!p){np -> fa = SAM_root;return;}SAM_Node *q = p -> next[x];if (q -> len == p-> len+1){np -> fa = q;return;}SAM_Node *nq = newSAM_Node(q);nq -> len = p -> len + 1;q -> fa = nq;np -> fa = nq;for (;p && p-> next[x] == q; p = p -> fa)p -> next[x] = nq;}int topcnt[MAXN];SAM_Node *topsam[MAXN*2];void SAM_build(char *s){SAM_init();int len = strlen(s);for (int i = 0 ;i < len; ++ i)SAM_add(s[i] - 'a' , i + 1);}void SAM_search(){SAM_Node *now = SAM_root;int len = strlen(str);int ans = 0;int maxans =0;for (int i = 0; i < len; ++ i){int t = str[i] - 'a';if (now -> next[t]){now =now -> next[t];++ans;}else{while (now != SAM_root && !now -> next[t])now = now -> fa;if (now -> next[t]){ans = now -> len + 1;now = now -> next[t];}else{now = SAM_root;ans = 0;}}maxans = max(maxans, ans);}cout << maxans << endl;}int r[MAXN], f[MAXN];void doit(){int n = strlen(str);SAM_build(str);/*for (int i = 0; i < SAM_size; ++ i){cout <<"@"<< SAM_node[i].id<< endl;for (int j = 0; j < 26; ++ j){if (SAM_node[i].next[j]){cout << SAM_node[i].next[j] -> id << endl;}}if (i)cout << "fa: " << SAM_node[i].fa -> id << endl;cout<<"===="<<endl;}exit(0);*/memset(topcnt, 0, sizeof(topcnt));memset(f, 0, sizeof(f));memset(r, 0, sizeof(r));for (int i = 0; i < SAM_size; ++ i)topcnt[SAM_node[i].len] ++ ;for (int i = 1; i <= n; ++ i)topcnt[i] += topcnt[i - 1];for (int i = 0; i < SAM_size ; ++ i)topsam[-- topcnt[SAM_node[i].len]] = &SAM_node[i];//topsam[i]第i个是谁，满足拓扑序SAM_Node *now = SAM_root;//cout<<"@"<<endl;for (int i = 0; i < n; ++ i){int t = str[i] - 'a';now = now -> next[t];r[now -> id] = 1;}//cout<<"!"<<endl;for (int i = SAM_size -1; i > 0; -- i){SAM_Node* p = topsam[i];//cout<< p->id <<endl;//cout<<"!!"<< p->fa->id << endl;r[(p->fa)->id] += r[p->id];}//cout<<"x"<<endl;for (int i = 1; i < SAM_size; ++ i){f[SAM_node[i].len] = max(f[SAM_node[i].len] , r[i]);}for (int i = n; i >= 0; -- i)f[i] = max(f[i + 1], f[i]);for (int i = 1; i <= n; ++ i)printf("%d\n", f[i]);}int main(){int t;while (~scanf("%s", str)){doit();}return 0;}