Searching the String zju3228

通过这题改正了以前一个重复模板匹配很低效的地方,见注释

#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;typedef long long LL;typedef unsigned long long ULL;typedef pair<int, int> PAIR;typedef multimap<int, int> MMAP;const int MAXN(600010);const int SIGMA_SIZE(26);const int MAXM(110);const int MAXE(4000010);const int MAXH(18);const int INFI((INT_MAX-1) >> 2);const int BASE(131);const int MOD(20090717);const ULL LIM(1000000000000000ull);char str[100010];int point[100010], type[100010];  //每个询问串所指向的节点，每个询问的类型int pos[MAXN];int at[2][MAXN];struct Q{int ind, next;};struct AC{int ch[MAXN][SIGMA_SIZE];int val[MAXN], f[MAXN], last[MAXN];int size;void init(){memset(ch[0], 0, sizeof(ch[0]));f[0] = val[0] = last[0] = 0;at[0][0] = at[1][0] = 0;pos[0] = -1;size = 1;}inline int idx(char temp){return temp-'a';}void insert(char *S, int tv, int len){int u = 0, id;for(; *S; ++S){id = idx(*S);if(!ch[u][id]){memset(ch[size], 0, sizeof(ch[size]));val[size] = 0;at[0][size] = at[1][size] = 0;pos[size] = -1;ch[u][id] = size++;}u = ch[u][id];}val[u] = len;  point[tv] = u;  //处理重复模板时一定不要用vector或者链式存储每个询问//而是每个询问建立一个指针指向该节点}int que[MAXN];int front, back;void construct(){front = back = 0;int cur, u;for(int i = 0; i < SIGMA_SIZE; ++i){u = ch[0][i];if(u){que[back++] = u;f[u] = last[u] = 0;}}while(front < back){cur = que[front++];for(int i = 0; i < SIGMA_SIZE; ++i){u = ch[cur][i];if(u){que[back++] = u;f[u] = ch[f[cur]][i];last[u] = val[f[u]]? f[u]: last[f[u]];}elsech[cur][i] = ch[f[cur]][i];}}}void find(char *T){int u = 0, id, ti = 0;for(; *T; ++T, ++ti){id = idx(*T);u = ch[u][id];for(int pi = u; pi; pi = last[pi]){//如果采用vector或链式解决重复模板的话，当重复的串较多时//效率将会受很大影响，采用询问指针来解决则没有这个问题//因为采用指针解决，重复模板对AC自动机来说是透明的，自动机认为只有一个模板串//而指针可以解释该节点属于哪个模板串++at[0][pi];if(ti-pos[pi] >= val[pi]){++at[1][pi];pos[pi] = ti;}}}}};AC ac;void solve(int n){ac.find(str);for(int i = 1; i <= n; ++i)printf("%d\n", at[type[i]][point[i]]);printf("\n");}int main(){char tstr[10];int n_case(0);while(~scanf("%s", str)){int n;scanf("%d", &n);ac.init();for(int i = 1; i <= n; ++i){scanf("%d%s", type+i, tstr);ac.insert(tstr, i, strlen(tstr));}ac.construct();printf("Case %d\n", ++n_case);solve(n);}return 0;}