[csu/coj 1632]LCP

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 }