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。
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 |
#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=0;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?1:-1; 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?1:-1;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=0;i<b.size();i++)a[i]=b[i];} template<typename T> void A2V(vector<T>&a,const T b[]){for(int i=0;i<a.size();i++)a[i]=b[i];} const double PI = acos(-1.0); const int INF = 1e9 + 7; /* -------------------------------------------------------------------------------- */ const int maxn = 2e3 + 7; 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 = 0; i < n; i ++) this->s[i] = s[i]; this->s[n] = 0; this->n = n + 1; this->m = m; } void build() { int i, *x = t, *y = t2; for (i = 0; i < m; i ++) c[i] = 0; for (i = 0; i < n; i ++) c[x[i] = s[i]] ++; for (i = 0; i < m; i ++) c[i] += c[i - 1]; for (i = n - 1; i >= 0; i --) sa[-- c[x[i]]] = i; for (int k = 1; k <= n; k <<= 1) { int p = 0; for (i = n - k; i < n; i ++) y[p ++] = i; for (i = 0; i < n; i ++) if (sa[i] >= k) y[p ++] = sa[i] - k; for (i = 0; i < m; i ++) c[i] = 0; for (i = 0; i < n; i ++) c[x[y[i]]] ++; for (i = 0; i < m; i ++) c[i] += c[i - 1]; for (i = n - 1; i >= 0; i --) sa[-- c[x[y[i]]]] = y[i]; swap(x, y); p = 1; x[sa[0]] = 0; for (i = 1; i < n; i ++) { x[sa[i]] = y[sa[i - 1]] == y[sa[i]] && y[sa[i - 1] + k] == y[sa[i] + k]? p - 1 : p ++; } if (p >= n) break; m = p; } } void getHeight() { int i, k = 0; for (i = 0; i < n; i ++) Rank[sa[i]] = i; for (i = 0; i < n; i ++) { if (k) k --; int j = sa[Rank[i] - 1]; 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 = 0; s[i]; i ++) { sa.init(i + 1, 128, 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 - 1], *ps = S[v - 1], common = 0, ans = 0; u --; for (int i = 1; i <= v; i ++) { if (ps[i] >= u) { ans += v - ps[i] - common; common = INF; } umin(common, ph[i + 1]); } printf("%d ", ans); } } return 0; } |