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 2333 using 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(); }