UVa 10829 L-Gap Substrings (后缀数组+rmq)

题意：给定上一个串，问你多少种UVU这一种形式的串，其中U不为空并且V的长度给定了。

析：枚举 U 的长度L，那么U一定是经过 0 L 2L 3L .... 其中的一个，所以求两个长度反lcp，一个向前延伸lcp1，一个向后延伸lcp2，然后加起来，要保证每个都不超过L，否则就会重复，然后个数就是 lcp1+lcp2-L。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define pu push_up
#define pd push_down
#define cl clear()
#define all 1,n,1
#define FOR(i,x,n)  for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e5 + 10;
const int maxm = 1e6 + 10;
const int mod = 50007;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, -1, 0, 1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
  return r >= 0 && r < n && c >= 0 && c < m;
}
struct Suffix_array{
  int s[maxn], sa[maxn], t[maxn], t2[maxn];
  int c[maxn], h[maxn], r[maxn];
  int n;
  int dp[maxn][20];
  void init(){ n = 0;  ms(sa, 0); }

  void build_sa(int m){
    int *x = t, *y = t2;
    for(int i = 0; i < m; ++i)  c[i] = 0;
    for(int i = 0; i < n; ++i)  ++c[x[i] = s[i]];
    for(int i = 1; i < m; ++i)  c[i] += c[i-1];
    for(int i = n-1; i >= 0; --i)  sa[--c[x[i]]] = i;

    for(int k = 1; k <= n; k <<= 1){
      int p = 0;
      for(int i = n-k; i < n; ++i)  y[p++] = i;
      for(int i = 0; i < n; ++i)  if(sa[i] >= k)  y[p++] = sa[i] - k;
      for(int i = 0; i < m; ++i)  c[i] = 0;
      for(int i = 0; i < n; ++i)  ++c[x[y[i]]];
      for(int i = 1; i < m; ++i)  c[i] += c[i-1];
      for(int i = n-1; i >= 0; --i)  sa[--c[x[y[i]]]] = y[i];

      swap(x, y);
      p = 1;  x[sa[0]] = 0;
      for(int i = 1; i < n; ++i)
        x[sa[i]] = y[sa[i]] == y[sa[i-1]] && y[sa[i]+k] == y[sa[i-1]+k] ? p - 1 : p++;
      if(p >= n)  break;
      m = p;
    }
  }

  void getHight(){
    int k = 0;
    for(int i = 0; i < n; ++i)  r[sa[i]] = i;
    for(int i = 0; i < n; ++i){
      if(k)  --k;
      int j = sa[r[i]-1];
      while(s[i+k] == s[j+k])  ++k;
      h[r[i]] = k;
    }
  }

  void rmq_init(){
    for(int i = 1; i <= n; ++i)  dp[i][0] = h[i];
    for(int j = 1; (1<<j) <= n; ++j)
      for(int i = 1; i + (1<<j) <= n; ++i)
        dp[i][j] = min(dp[i][j-1], dp[i+(1<<j-1)][j-1]);
  }

  int query(int L, int R){
    L = r[L];  R = r[R];
    if(L > R)  swap(L, R);
    ++L;
    int k = log(R - L + 1.0) / log(2.0);
    return min(dp[L][k], dp[R-(1<<k)+1][k]);
  }

  LL solve(int len, int g){
    LL ans = 0;
    for(int i = 1; i < len; ++i){
      for(int j = 0; j + i + g < len; j += i){
        int lcp = min(i, query(j, j+i+g)) + min(query(n-1-j, n-1-j-i-g), i-1);
        ans += max(0, lcp - i + 1);
      }
    }
    return ans;
  }
};

Suffix_array arr;
char s[maxn];

int main(){
  int T;  cin >> T;
  for(int kase = 1; kase <= T; ++kase){
    scanf("%d", &n);  arr.init();
    scanf("%s", s);
    int len = strlen(s);
    for(int i = 0; i < len; ++i)  arr.s[arr.n++] = s[i] - 'a' + 1;
    arr.s[arr.n++] = 27;
    for(int i = len-1; i >= 0; --i)  arr.s[arr.n++] = s[i] - 'a' + 1;
    arr.s[arr.n++] = 0;
    arr.build_sa(30);
    arr.getHight();
    arr.rmq_init();
    printf("Case %d: %lld
", kase, arr.solve(len, n));
  }
  return 0;
}