[hdu4622 Reincarnation]后缀数组

题意：给一个长度为2000的字符串，10000次询问区间[L,R]内的不同子串的个数

思路：对原串的每个前缀求一边后缀数组，询问[L,R]就变成了询问[L,n]了，即求一个后缀里面出现了多少个不同子串。于是对所有大于等于L的后缀统计一遍即可。

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#include <map> #include <set> #include <cmath> #include <ctime> #include <deque> #include <queue> #include <stack> #include <vector> #include <cstdio> #include <string> #include <cstdlib> #include <cstring> #include <iostream> #include <algorithm> using namespace std; #define X first #define Y second #define pb push_back #define mp make_pair #define all(a) (a).begin(), (a).end() #define fillchar(a, x) memset(a, x, sizeof(a)) #define copy(a, b) memcpy(a, b, sizeof(a)) typedef long long ll; typedef pair<int, int> pii; typedef unsigned long long ull; #ifndef ONLINE_JUDGE void RI(vector<int>&a,int n){a.resize(n);for(int i=;i<n;i++)scanf("%d",&a[i]);} void RI(){}void RI(int&X){scanf("%d",&X);}template<typename...R> void RI(int&f,R&...r){RI(f);RI(r...);}void RI(int*p,int*q){int d=p<q?:-; while(p!=q){scanf("%d",p);p+=d;}}void print(){cout<<endl;}template<typename T> void print(const T t){cout<<t<<endl;}template<typename F,typename...R> void print(const F f,const R...r){cout<<f<<", ";print(r...);}template<typename T> void print(T*p, T*q){int d=p<q?:-;while(p!=q){cout<<*p<<", ";p+=d;}cout<<endl;} #endif template<typename T>bool umax(T&a, const T&b){return b<=a?false:(a=b,true);} template<typename T>bool umin(T&a, const T&b){return b>=a?false:(a=b,true);} template<typename T> void V2A(T a[],const vector<T>&b){for(int i=;i<b.size();i++)a[i]=b[i];} template<typename T> void A2V(vector<T>&a,const T b[]){for(int i=;i<a.size();i++)a[i]=b[i];} const double PI = acos(-1.0); const int INF = 1e9 + ; /* -------------------------------------------------------------------------------- */ const int maxn = 2e3 + ; struct SA { //const static int maxn = 2e3 + 7; int sa[maxn], t[maxn], t2[maxn], c[maxn], n, m; int Rank[maxn], Height[maxn]; int s[maxn]; void init(int n, int m, char s[]) { for (int i = ; i < n; i ++) this->s[i] = s[i]; this->s[n] = ; this->n = n + ; this->m = m; } void build() { int i, *x = t, *y = t2; for (i = ; i < m; i ++) c[i] = ; for (i = ; i < n; i ++) c[x[i] = s[i]] ++; for (i = ; i < m; i ++) c[i] += c[i - ]; for (i = n - ; i >= ; i --) sa[-- c[x[i]]] = i; for (int k = ; k <= n; k <<= ) { int p = ; for (i = n - k; i < n; i ++) y[p ++] = i; for (i = ; i < n; i ++) if (sa[i] >= k) y[p ++] = sa[i] - k; for (i = ; i < m; i ++) c[i] = ; for (i = ; i < n; i ++) c[x[y[i]]] ++; for (i = ; i < m; i ++) c[i] += c[i - ]; for (i = n - ; i >= ; i --) sa[-- c[x[y[i]]]] = y[i]; swap(x, y); p = ; x[sa[]] = ; for (i = ; i < n; i ++) { x[sa[i]] = y[sa[i - ]] == y[sa[i]] && y[sa[i - ] + k] == y[sa[i] + k]? p - : p ++; } if (p >= n) break; m = p; } } void getHeight() { int i, k = ; for (i = ; i < n; i ++) Rank[sa[i]] = i; for (i = ; i < n; i ++) { if (k) k --; int j = sa[Rank[i] - ]; while (s[i + k] == s[j + k]) k ++; Height[Rank[i]] = k; } } }; SA sa; int H[maxn][maxn], S[maxn][maxn]; char s[maxn]; int main() { #ifndef ONLINE_JUDGE freopen("in.txt", "r", stdin); //freopen("out.txt", "w", stdout); #endif // ONLINE_JUDGE int T, m; cin >> T; while (T --) { scanf("%s", s); for (int i = ; s[i]; i ++) { sa.init(i + , , s); sa.build(); sa.getHeight(); copy(H[i], sa.Height); copy(S[i], sa.sa); } cin >> m; while (m --) { int u, v; scanf("%d%d", &u, &v); int *ph = H[v - ], *ps = S[v - ], common = , ans = ; u --; for (int i = ; i <= v; i ++) { if (ps[i] >= u) { ans += v - ps[i] - common; common = INF; } umin(common, ph[i + ]); } printf("%d\n", ans); } } return ; }