HDU

HDU - 4622 Reincarnation

给定字符串S，有q次查询，每次查询[l, r]内的子串的数量。

#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <stack>#include <set>#include <map>#include <string>#include <math.h>#include <stdlib.h>#include <time.h>#include <bitset>#define INF 0x3f3f3f3f#define eps 1e-6#define PI 3.1415926#define mod 1000000009#define base 2333using namespace std;typedef long long LL;const int maxn = 2e3 + 10;const int maxx = 1e3 + 10;inline void splay(int &v) {    v=0;char c=0;int p=1;    while(c<'0' || c >'9'){if(c=='-')p=-1;c=getchar();}    while(c>='0' && c<='9'){v=(v<<3)+(v<<1)+c-'0';c=getchar();}    v*=p;}struct Node {    Node *pre, *nxt[26];    int step;    void Clear() {        step = 0;        pre = NULL;        memset(nxt, NULL, sizeof(nxt));    }    int calc() {        if(pre == NULL) return 0;        return step - pre->step;    }} *root, *last;Node st[maxn<<1], *cur;char str[maxn];int t, q, l, r, cnt, ans[maxn][maxn];void init() {    cnt = 0;    cur = st;    root = last = cur++;    root->Clear();}void extend(int w) {    Node *np = cur++, *p = last;    np->Clear();    np->step = p->step+1;    while(p && !p->nxt[w])        p->nxt[w] = np, p = p->pre;    if(p == NULL) {        np->pre = root;        cnt += np->calc();    }    else {        Node *q = p->nxt[w];        if(q->step == p->step+1) {            np->pre = q;            cnt += np->calc();        }        else {            Node *nq = cur++;            nq->Clear();            memcpy(nq->nxt, q->nxt, sizeof(q->nxt));            cnt -= q->calc();            nq->step = p->step+1;            nq->pre = q->pre;            np->pre = q->pre = nq;            cnt += q->calc()+np->calc()+nq->calc();            while(p && p->nxt[w] == q)                p->nxt[w] = nq, p = p->pre;        }    }    last = np;}void solve() {    scanf("%d", &t);    while(t--) {        scanf("%s", str);        int len = strlen(str);        for(int i = 0; i < len; i++) {            init();            for(int j = i; j < len; j++) {                extend(str[j]-'a');                ans[i][j] = cnt;            }        }        splay(q);        while(q--) {            splay(l), splay(r);            printf("%d\n", ans[l-1][r-1]);        }    }}int main() {    //srand(time(NULL));    //freopen("kingdom.in","r",stdin);    //freopen("kingdom.out","w",stdout);    solve();}