str2int hdoj4436

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;}