由于是处理多个串的子串问题,所以需要先把这些串连接起来,但还要防止连接起来之后子串跨越多个串,所以每个子串之间用一个没有出现的字符分隔,和后缀数组稍有不同,这题的分隔符可以都相同,因为只要转移的过程当中不通过分隔符转移即可。每个节点要额外记录俩个信息,一个是从根节点到达当前节点的子串个数,另一个是从根节点到达当前节点所生成的串前缀和取模,然后按照len从小到大的拓扑序递推即可。做完这题发现了一件蛋疼的事情,那就是发现今天竟然已经20号了,嘛,我们24号就考试了啊,一点还没复习呢,泪奔滚去复习了。
#include <iostream> #include <cstdio> #include <cstdlib> #include <cmath> #include <queue> #include <algorithm> #include <vector> #include <cstring> #include <stack> #include <cctype> #include <utility> #include <map> #include <string> #include <climits> #include <set> #include <string> #include <sstream> #include <utility> #include <ctime> using std::priority_queue; using std::vector; using std::swap; using std::stack; using std::sort; using std::max; using std::min; using std::pair; using std::map; using std::string; using std::cin; using std::cout; using std::set; using std::queue; using std::string; using std::istringstream; using std::make_pair; using std::getline; using std::greater; using std::endl; using std::multimap; using std::deque; typedef long long LL; typedef unsigned long long ULL; typedef pair<int, int> PAIR; typedef multimap<int, int> MMAP; const int MAXN(111000); const int SIGMA_SIZE(11); const int MAXM(110); const int MAXE(300010); const int MAXH(18); const int INFI((INT_MAX-1) >> 1); const int MOD(2012); const ULL BASE(31); const ULL LIM(1000000000000000ull); inline int idx(char temp) { return temp-'0'; } char str[MAXN]; int cnt[MAXN]; int buc[MAXN << 1]; struct SAM { struct NODE { int len; int count1, count2; //count1是从根节点到达当前节点的子串个数,count2从根节点到达当前节点所生成的串前缀和取模 NODE *f, *ch[SIGMA_SIZE]; }; NODE *root, *last; NODE pool[MAXN << 1]; int size; void init() { root = last = pool; root->f = 0; root->len = 0; root->count1 = root->count2 = 0; memset(root->ch, 0, sizeof(root->ch)); size = 1; } NODE *newnode(int tl) { pool[size].len = tl; pool[size].count1 = pool[size].count2 = 0; memset(pool[size].ch, 0, sizeof(pool[size].ch)); return pool+size++; } void extend(int id) { NODE *p = last, *np = newnode(last->len+1); last = np; while(p && p->ch[id] == 0) p->ch[id] = np, p = p->f; if(p == 0) np->f = root; else { NODE *q = p->ch[id]; if(p->len+1 == q->len) np->f = q; else { NODE *nq = newnode(p->len+1); memcpy(nq->ch, q->ch, sizeof(nq->ch)); nq->f = q->f; q->f = np->f = nq; while(p && p->ch[id] == q) p->ch[id] = nq, p = p->f; } } } /* SAM::NODE *que[MAXN << 1]; int front, back; void getMin() //得到每个状态的min { front = back = 0; que[back++] = root; SAM::NODE *p; while(front < back) { p = que[front++]; for(int i = 0; i < SIGMA_SIZE; ++i) if(p->ch[i] && !p->ch[i]->mi) { p->ch[i]->mi = p->mi+1; que[back++] = p->ch[i]; } } } */ void turpo() //按MAX拓扑 { memset(cnt, 0, sizeof(cnt[0])*(last->len+1)); for(int i = 0; i < size; ++i) ++cnt[pool[i].len]; for(int i = 1; i <= last->len; ++i) cnt[i] += cnt[i-1]; for(int i = 0; i < size; ++i) buc[--cnt[pool[i].len]] = i; } /* void getRight() //得到每个状态的right { NODE *tp = root; for(char *sp = str; *sp; ++sp) tp = tp->ch[idx(*sp)], tp->right = 1; for(int i = size-1; i >= 0; --i) { tp = pool+buc[i]; if(tp->f) tp->f->right += tp->right; } }*/ }; SAM sam; int main() { int n; while(~scanf("%d", &n)) { sam.init(); for(int i = 0; i < n; ++i) { if(i) sam.extend(10); scanf("%s", str); for(char *sp = str; *sp; ++sp) sam.extend(idx(*sp)); } sam.turpo(); sam.root->count1 = 1; for(int i = 1; i < 10; ++i) if(sam.root->ch[i]) { sam.root->ch[i]->count1 += sam.root->count1; sam.root->ch[i]->count2 = (sam.root->ch[i]->count2+sam.root->count2*10+sam.root->count1*i)%MOD; } SAM::NODE *p; int ans = 0; for(int i = 1; i < sam.size; ++i) { p = sam.pool+buc[i]; ans = (ans+p->count2)%MOD; for(int j = 0; j < 10; ++j) if(p->ch[j]) { p->ch[j]->count1 += p->count1; p->ch[j]->count2 = (p->ch[j]->count2+p->count2*10+p->count1*j)%MOD; } } printf("%d\n", ans); } return 0; }