标签:
题意:求一个串的出现次数超过1次的字串的个数
思路:对于一个后缀,出现在它后面的所有后缀与它的LCP的最大值就是应该增加的答案,当然这里没有考虑去重,但是却转化了问题,使得我们可以用最长公共前缀来统计答案。假设我们将每一个后缀按字典序排好,那么对于每一个后缀,与其它后缀的LCP的最大值其实就是与它相邻的两个的lcp的较大值,这不就是height数组了么?考虑去重的问题,如果height[i]>height[i-1],那么对于rank为i和i-1的最长公共前缀,它的前height[i-1]个前缀已经统计过了,答案只需要加上height[i]-height[i-1],如果height[i]<=height[i-1]则不需处理。
1 #pragma comment(linker, "/STACK:10240000,10240000") 2 3 #include <iostream> 4 #include <cstdio> 5 #include <algorithm> 6 #include <cstdlib> 7 #include <cstring> 8 #include <map> 9 #include <queue> 10 #include <deque> 11 #include <cmath> 12 #include <vector> 13 #include <ctime> 14 #include <cctype> 15 #include <set> 16 #include <bitset> 17 #include <functional> 18 #include <numeric> 19 #include <stdexcept> 20 #include <utility> 21 22 using namespace std; 23 24 #define mem0(a) memset(a, 0, sizeof(a)) 25 #define mem_1(a) memset(a, -1, sizeof(a)) 26 #define lson l, m, rt << 1 27 #define rson m + 1, r, rt << 1 | 1 28 #define rep_up0(a, b) for (int a = 0; a < (b); a++) 29 #define rep_up1(a, b) for (int a = 1; a <= (b); a++) 30 #define rep_down0(a, b) for (int a = b - 1; a >= 0; a--) 31 #define rep_down1(a, b) for (int a = b; a > 0; a--) 32 #define all(a) (a).begin(), (a).end() 33 #define lowbit(x) ((x) & (-(x))) 34 #define constructInt4(name, a, b, c, d) name(int a = 0, int b = 0, int c = 0, int d = 0): a(a), b(b), c(c), d(d) {} 35 #define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {} 36 #define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {} 37 #define pchr(a) putchar(a) 38 #define pstr(a) printf("%s", a) 39 #define sstr(a) scanf("%s", a) 40 #define sint(a) scanf("%d", &a) 41 #define sint2(a, b) scanf("%d%d", &a, &b) 42 #define sint3(a, b, c) scanf("%d%d%d", &a, &b, &c) 43 #define pint(a) printf("%d\n", a) 44 #define test_print1(a) cout << "var1 = " << a << endl 45 #define test_print2(a, b) cout << "var1 = " << a << ", var2 = " << b << endl 46 #define test_print3(a, b, c) cout << "var1 = " << a << ", var2 = " << b << ", var3 = " << c << endl 47 #define mp(a, b) make_pair(a, b) 48 #define pb(a) push_back(a) 49 50 typedef unsigned int uint; 51 typedef long long LL; 52 typedef pair<int, int> pii; 53 typedef vector<int> vi; 54 55 const int dx[8] = {0, 0, -1, 1, 1, 1, -1, -1}; 56 const int dy[8] = {-1, 1, 0, 0, 1, -1, 1, -1 }; 57 const int maxn = 120000; 58 const int md = 1000000007; 59 const int inf = 1e9 + 7; 60 const LL inf_L = 1e18 + 7; 61 const double pi = acos(-1.0); 62 const double eps = 1e-6; 63 64 template<class T>T gcd(T a, T b){return b==0?a:gcd(b,a%b);} 65 template<class T>bool max_update(T &a,const T &b){if(b>a){a = b; return true;}return false;} 66 template<class T>bool min_update(T &a,const T &b){if(b<a){a = b; return true;}return false;} 67 template<class T>T condition(bool f, T a, T b){return f?a:b;} 68 template<class T>void copy_arr(T a[], T b[], int n){rep_up0(i,n)a[i]=b[i];} 69 int make_id(int x, int y, int n) { return x * n + y; } 70 71 /// 构造后缀数组的之前,需要在原串末尾加个空字符(比其它字符都小即可), 72 ///把这个空字符看成原串的一部分(这样在比较的时候到这个位置一定可以分个大小), 73 ///所以n应该为原序列长度+1,后缀n-1是"空串",sa[0]总是n-1。 74 struct SuffixArray { 75 int n; 76 int arr[6][maxn]; 77 int *sa, *x, *y, *c, *rnk, *height; 78 SuffixArray() { sa = arr[0]; x = arr[1]; y = arr[2]; c = arr[3]; rnk = arr[4]; height = arr[5]; } 79 void resize(int nn) { n = nn; mem0(arr[0]); } 80 void build_sa(int s[], int m) { // m is biger than the max value of char 81 rep_up0(i, m) c[i] = 0; 82 rep_up0(i, n) c[x[i] = s[i]]++; 83 rep_up1(i, m - 1) c[i] += c[i - 1]; 84 rep_down0(i, n) sa[--c[x[i]]] = i; 85 for (int k = 1; k <= n; k <<= 1) { 86 int p = 0; 87 for (int i = n - k; i < n; i++) y[p++] = i; 88 rep_up0(i, n) if (sa[i] >= k) y[p++] = sa[i] - k; 89 rep_up0(i, m) c[i] = 0; 90 rep_up0(i, n) c[x[y[i]]]++; 91 rep_up0(i, m) c[i] += c[i - 1]; 92 rep_down0(i, n) sa[--c[x[y[i]]]] = y[i]; 93 swap(x, y); 94 p = 1; x[sa[0]] = 0; 95 for (int i = 1; i < n; i++) { 96 x[sa[i]] = y[sa[i - 1]] == y[sa[i]] && y[sa[i - 1] + k] == y[sa[i] + k]? p - 1 : p++; 97 } 98 if (p >= n) break; 99 m = p; 100 } 101 } 102 void build_height(int s[]) { 103 mem0(height); 104 int k = 0; 105 rep_up0(i, n) rnk[sa[i]] = i; 106 rep_up0(i, n) { 107 if (k) k--; 108 int j = sa[rnk[i] - 1]; 109 while (s[i + k] == s[j + k]) k++; 110 height[rnk[i]] = k; 111 } 112 } 113 int solve(int n) { 114 int ans = 0; 115 for (int i = 2; i <= n; i ++) { 116 if (height[i] > height[i - 1]) ans += height[i] - height[i - 1]; 117 } 118 return ans; 119 } 120 }; 121 char s[maxn]; 122 int ss[maxn]; 123 SuffixArray sa; 124 int main() { 125 //freopen("in.txt", "r", stdin); 126 int T; 127 cin >> T; 128 while (T --) { 129 scanf("%s", s); 130 int len = strlen(s) + 1; 131 rep_up0(i, len) ss[i] = s[i]; 132 sa.resize(len); 133 sa.build_sa(ss, 128); 134 sa.build_height(ss); 135 cout << sa.solve(len - 1) << endl; 136 } 137 return 0; 138 }
标签:
原文地址:http://www.cnblogs.com/jklongint/p/4526506.html