题目大意:
就是给出一个字符串(长度不超过10W), 接着Q次询问, 对于每次询问要找到第K小子串, 注意这里的字串是本质不同的子串,
大致思路:
由于Q可以达到10^5 使用后缀自动机的话容易超时, 于是需要使用后缀数组, 对于给出的字符串求出后缀数组之后利用height数组处理出前i个分支下有多少个不同的子串, 那么对于每次询问二分查找即可, 由于每次回答的[l, r]中l药最小, 二分之后向后再找一点距离即可
代码如下:
Result : Accepted Memory : 5488 KB Time : 592 ms
/* * Author: Gatevin * Created Time: 2015/5/4 15:22:39 * File Name: Rin_Tohsaka.cpp */ #include<iostream> #include<sstream> #include<fstream> #include<vector> #include<list> #include<deque> #include<queue> #include<stack> #include<map> #include<set> #include<bitset> #include<algorithm> #include<cstdio> #include<cstdlib> #include<cstring> #include<cctype> #include<cmath> #include<ctime> #include<iomanip> using namespace std; const double eps(1e-8); typedef long long lint; #define foreach(e, x) for(__typeof(x.begin()) e = x.begin(); e != x.end(); ++e) #define SHOW_MEMORY(x) cout<<sizeof(x)/(1024*1024.)<<"MB"<<endl #define rank rankrankrank #define maxn 100010 int wa[maxn], wb[maxn], wv[maxn], Ws[maxn]; int cmp(int *r, int a, int b, int l) { return r[a] == r[b] && r[a + l] == r[b + l]; } void da(int *r, int *sa, int n, int m) { int *x = wa, *y = wb, *t, i, j, p; for(i = 0; i < m; i++) Ws[i] = 0; for(i = 0; i < n; i++) Ws[x[i] = r[i]]++; for(i = 1; i < m; i++) Ws[i] += Ws[i - 1]; for(i = n - 1; i >= 0; i--) sa[--Ws[x[i]]] = i; for(j = 1, p = 1; p < n; j <<= 1, m = p) { for(p = 0, i = n - j; i < n; i++) y[p++] = i; for(i = 0; i < n; i++) if(sa[i] >= j) y[p++] = sa[i] - j; for(i = 0; i < n; i++) wv[i] = x[y[i]]; for(i = 0; i < m; i++) Ws[i] = 0; for(i = 0; i < n; i++) Ws[wv[i]]++; for(i = 1; i < m; i++) Ws[i] += Ws[i - 1]; for(i = n - 1; i >= 0; i--) sa[--Ws[wv[i]]] = y[i]; for(t = x, x = y, y = t, p = 1, x[sa[0]] = 0, i = 1; i < n; i++) x[sa[i]] = cmp(y, sa[i - 1], sa[i], j) ? p - 1 : p++; } return; } int rank[maxn], height[maxn]; void calheight(int *r, int *sa, int n) { int i, j, k = 0; for(i = 1; i <= n; i++) rank[sa[i]] = i; for(i = 0; i < n; height[rank[i++]] = k) for(k ? k-- : 0, j = sa[rank[i] - 1]; r[i + k] == r[j + k]; k++); return; } char in[maxn]; int s[maxn], sa[maxn]; lint cnt[maxn]; lint v, l, r, k; void init(int len) { cnt[0] = 0; for(int i = 1; i <= len; i++) cnt[i] = cnt[i - 1] + (len - sa[i]) - height[i]; return; } void solve(int len)//找出第k大的位置 { if(k > cnt[len]) { l = r = 0; return; } int L = 0, R = len, mid, res; while(L <= R) { mid = (L + R) >> 1; if(cnt[mid] < k) { res = mid; L = mid + 1; } else R = mid - 1; } int ad = k - cnt[res]; int LEN = ad + height[res + 1];//这个子串的长度 l = sa[res + 1]; res += 2; while(height[res] >= LEN)//向后找使得l最小 l = min(l, (lint)sa[res]), res++; r = l + LEN - 1; l++, r++; } int main() { while(scanf("%s", in) != EOF) { int len = strlen(in); for(int i = 0; i < len; i++) s[i] = in[i] - 'a' + 1; s[len] = 0; da(s, sa, len + 1, 28); calheight(s, sa, len); init(len); int Q; scanf("%d", &Q); l = r = 0; while(Q--) { scanf("%I64d", &v); k = (l^r^v) + 1; solve(len); printf("%I64d %I64d\n", l, r); } } return 0; }