[hdu4552]最长公共前缀

题意：给一个串s，求s的每个前缀出现次数之和。

思路：对于一个后缀i，设i和原串的最长公共前缀为k，则当前总共可以产生k个答案。因此原题转化为求所有后缀与原串的最长公共前缀之和。模板题。以下为通过模板。

 #pragma comment(linker, "/STACK:10240000,10240000")

 #include <iostream>
 #include <cstdio>
 #include <algorithm>
 #include <cstdlib>
 #include <cstring>
 #include <map>
 #include <queue>
 #include <deque>
 #include <cmath>
 #include <vector>
 #include <ctime>
 #include <cctype>
 #include <set>
 #include <bitset>
 #include <functional>
 #include <numeric>
 #include <stdexcept>
 #include <utility>

 using namespace std;

 #define mem0(a) memset(a, 0, sizeof(a))
 #define mem_1(a) memset(a, -1, sizeof(a))
 #define lson l, m, rt << 1
 #define rson m + 1, r, rt << 1 | 1
 #define define_m int m = (l + r) >> 1
 #define rep_up0(a, b) for (int a = 0; a < (b); a++)
 #define rep_up1(a, b) for (int a = 1; a <= (b); a++)
 #define rep_down0(a, b) for (int a = b - 1; a >= 0; a--)
 #define rep_down1(a, b) for (int a = b; a > 0; a--)
 #define all(a) (a).begin(), (a).end()
 #define lowbit(x) ((x) & (-(x)))
 #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) {}
 #define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {}
 #define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {}
 #define pchr(a) putchar(a)
 #define pstr(a) printf("%s", a)
 #define sstr(a) scanf("%s", a)
 #define sint(a) scanf("%d", &a)
 #define sint2(a, b) scanf("%d%d", &a, &b)
 #define sint3(a, b, c) scanf("%d%d%d", &a, &b, &c)
 #define pint(a) printf("%d\n", a)
 #define test_print1(a) cout << "var1 = " << a << endl
 #define test_print2(a, b) cout << "var1 = " << a << ", var2 = " << b << endl
 #define test_print3(a, b, c) cout << "var1 = " << a << ", var2 = " << b << ", var3 = " << c << endl

 typedef double db;
 typedef long long LL;
 typedef pair<int, int> pii;
 typedef multiset<int> msi;
 typedef set<int> si;
 typedef vector<int> vi;
 typedef map<int, int> mii;

 const int dx[] = {, , -, , , , -, -};
 const int dy[] = {-, , , , , -, , - };
 const int maxn = 1e5 + ;
 const int md = 1e9 + ;
 const int inf = 1e9 + ;
 const LL inf_L = 1e18 + ;
 const double pi = acos(-1.0);
 const double eps = 1e-;

 template<class T>T gcd(T a, T b){return b==?a:gcd(b,a%b);}
 template<class T>bool max_update(T &a,const T &b){if(b>a){a = b; return true;}return false;}
 template<class T>bool min_update(T &a,const T &b){if(b<a){a = b; return true;}return false;}
 template<class T>T condition(bool f, T a, T b){return f?a:b;}
 template<class T>void copy_arr(T a[], T b[], int n){rep_up0(i,n)a[i]=b[i];}
 int make_id(int x, int y, int n) { return x * n + y; }

 struct SparseTable {
     int d[maxn][];
     int t[maxn];
     int n;
     void resize(int nn) { n = nn; }
     void Init_min(int a[]) {
         rep_up0(i, n) d[i][] = a[i];
         for (int j = ; ( << j) <= n; j++) {
             for (int i = ; i + ( << j) -  < n; i++) {
                 d[i][j] = min(d[i][j - ], d[i + ( << (j - ))][j - ]);
             }
         }
         int p = -;
         rep_up1(i, n) {
             if ((i & (i - )) == ) p++;
             t[i] = p;
         }
     }
     void Init_max(int a[]) {
         rep_up0(i, n) d[i][] = a[i];
         for (int j = ; ( << j) <= n; j++) {
             for (int i = ; i + ( << j) -  < n; i++) {
                 d[i][j] = max(d[i][j - ], d[i + ( << (j - ))][j - ]);
             }
         }
         int p = -;
         rep_up1(i, n) {
             if ((i & (i - ) == )) p++;
             t[i] = p;
         }
     }
     int RMQ_min(int L, int R) {
         int p = t[R - L + ];
         return min(d[L][p], d[R - ( << p) + ][p]);
     }
     int RMQ_max(int L, int R) {
         int p = t[R - L + ];
         return max(d[L][p], d[R - ( << p) + ][p]);
     }
 };

 /// 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)，
 ///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）,
 ///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。
 struct SuffixArray {
     int n;
     int arr[][maxn];
     int *sa = arr[], *x = arr[], *y = arr[], *c = arr[], *rnk = arr[], *height = arr[];
     void resize(int nn) {  n = nn; mem0(arr[]); }
     void build_sa(int s[], int m) { // m is biger than the max value of char
         rep_up0(i, m) c[i] = ;
         rep_up0(i, n) c[x[i] = s[i]]++;
         rep_up1(i, m - ) c[i] += c[i - ];
         rep_down0(i, n) sa[--c[x[i]]] = i;
         for (int k = ; k <= n; k <<= ) {
             int p = ;
             for (int i = n - k; i < n; i++) y[p++] = i;
             rep_up0(i, n) if (sa[i] >= k) y[p++] = sa[i] - k;
             rep_up0(i, m) c[i] = ;
             rep_up0(i, n) c[x[y[i]]]++;
             rep_up0(i, m) c[i] += c[i - ];
             rep_down0(i, n) sa[--c[x[y[i]]]] = y[i];
             swap(x, y);
             p = ; x[sa[]] = ;
             for (int 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 build_height(int s[]) {
         int k = ;
         rep_up0(i, n) rnk[sa[i]] = i;
         rep_up0(i, n) {
             if (k) k--;
             int j = sa[rnk[i] - ];
             while (s[i + k] == s[j + k]) k++;
             height[rnk[i]] = k;
         }
     }
 };

 SparseTable st;
 SuffixArray sa;
 char s[maxn];
 int ss[maxn];

 int main() {
     //freopen("in.txt", "r", stdin);
     while (~scanf("%s", s)) {
         int len = strlen(s) + ;
         rep_up0(i, len) ss[i] = s[i];
         sa.resize(len);
         sa.build_sa(ss, );
         sa.build_height(ss);
         st.resize(len);
         st.Init_min(sa.height);
         int ans = (len - ) % ;
         for (int i = ; i < len; i++) {
             int j = sa.rnk[], k = sa.rnk[i];
             if (j > k) swap(j, k);
             ans += st.RMQ_min(j + , k);
             ans %= ;
         }
         cout << ans << endl;
     }
     return ;
 }