后缀数组初步
2015年11月18日
uoj模板题：
1 #include <bits/stdc++.h> 2 #define rep(i, a, b) for (register int i = a; i <= b; i++) 3 #define drep(i, a, b) for (register int i = a; i >= b; i--) 4 #define REP(i, a, b) for (int i = a; i < b; i++) 5 #define pb push_back 6 #define mp make_pair 7 #define clr(x) memset(x, 0, sizeof(x)) 8 #define xx first 9 #define yy second 10 11 using namespace std; 12 13 typedef long long i64; 14 typedef pair<int, int> pii; 15 const int inf = ~0U >> 1; 16 const i64 INF = ~0ULL >> 1; 17 //******************************** 18 19 const int maxn = 100005; 20 21 char s[maxn]; 22 int sa[maxn], t[maxn], t2[maxn], c[maxn], n; 23 void build_sa(int m) { 24 int *x = t, *y = t2; 25 REP(i, 0, m) c[i] = 0; 26 REP(i, 0, n) c[x[i] = s[i]]++; 27 REP(i, 1, m) c[i] += c[i - 1]; 28 drep(i, n - 1, 0) sa[--c[x[i]]] = i; 29 for (int k = 1; k <= n; k <<= 1) { 30 int p = 0; 31 REP(i, n - k, n) y[p++] = i; 32 REP(i, 0 , n) if (sa[i] >= k) y[p++] = sa[i] - k; 33 34 REP(i, 0, m) c[i] = 0; 35 REP(i, 0, n) c[x[y[i]]]++; 36 REP(i, 1, m) c[i] += c[i - 1]; 37 drep(i, n - 1, 0) sa[--c[x[y[i]]]] = y[i]; 38 39 swap(x, y); 40 p = 1; x[sa[0]] = 0; 41 REP(i, 1, n) 42 x[sa[i]] = y[sa[i - 1]]==y[sa[i]] && y[sa[i - 1] + k]==y[sa[i] + k] ? p - 1 : p++; 43 if (p >= n) break; 44 m = p; 45 } 46 } 47 48 int rnk[maxn], height[maxn]; 49 void getHeight() { 50 REP(i, 0, n) rnk[sa[i]] = i; 51 int k(0); 52 REP(i, 0, n) { 53 if (k) k--; 54 int j = sa[rnk[i] - 1]; 55 while (s[i + k] == s[j + k]) k++; 56 height[rnk[i]] = k; 57 } 58 } 59 60 int main() { 61 scanf("%s", s); 62 n = strlen(s) + 1; 63 build_sa('z' + 1); 64 REP(i, 1, n) printf("%d ", sa[i] + 1); 65 puts(""); 66 getHeight(); 67 REP(i, 2, n) printf("%d ", height[i]); 68 return 0; 69 }
2015年11月20日
poj1743 <最大不重合公共子序列长度>
给出一个数列，差分之，求出最长差分相同的序列。
首先后缀a和后缀b的LCP为height[sa[a] + 1] - height[sa[b]]的最小值，假设sa[a] < sa[b];
那么若不考虑重合，在二分答案的同时，将height数组按大于等于二分值和小于二分值分段，比如(1, 2, 3, 4, 1, 2) 用2来二分 则划分为(1, (2, 3, 4), (1), (2));
再考虑重合，符合答案的情况则一定在大于二分值的每段中，只需找到每一个大于二分值的段中的最大sa和最小sa，然后Max - Min > mid 即可，>是因为差分后不能重合，若没有差分，则用>=
1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 #define rep(i, a, b) for (int i = a; i <= b; i++) 5 #define REP(i, a, b) for (int i = a; i < b; i++) 6 #define drep(i, a, b) for (int i = a; i >= b; i--) 7 #define pb push_back 8 #define mp make_pair 9 #define clr(x) memset(x, 0, sizeof(x)) 10 #define travel(x) for (int i = G[x]; i; i = e[i].nx) 11 #define xx first 12 #define yy second 13 14 using namespace std; 15 16 typedef long long i64; 17 typedef pair<int, int> pii; 18 const int inf = ~0U >> 1; 19 const i64 INF = ~0ULL >> 1; 20 //*************************** 21 22 const int maxn = 20005; 23 24 int s[maxn]; 25 int sa[maxn], t[maxn], t2[maxn], c[maxn], n; 26 void build_sa(int m) { 27 int *x = t, *y = t2; 28 REP(i, 0, m) c[i] = 0; 29 REP(i, 0, n) c[x[i] = s[i]]++; 30 REP(i, 1, m) c[i] += c[i - 1]; 31 drep(i, n - 1, 0) sa[--c[x[i]]] = i; 32 33 for (int k = 1; k <= n; k <<= 1) { 34 int p = 0; 35 REP(i, n - k, n) y[p++] = i; 36 REP(i, 0, n) if (sa[i] >= k) y[p++] = sa[i] - k; 37 38 REP(i, 0, m) c[i] = 0; 39 REP(i, 0, n) c[x[y[i]]]++; 40 REP(i, 1, m) c[i] += c[i - 1]; 41 drep(i, n - 1, 0) sa[--c[x[y[i]]]] = y[i]; 42 43 swap(x, y); 44 p = 1, x[sa[0]] = 0; 45 REP(i, 1, n) 46 x[sa[i]] = y[sa[i]] == y[sa[i - 1]] && y[sa[i] + k] == y[sa[i - 1] + k] ? p - 1 : p++; 47 if (p >= n) break; 48 m = p; 49 } 50 } 51 52 int rnk[maxn], height[maxn]; 53 void buildHeight() { 54 int k(0); 55 REP(i, 0, n) rnk[sa[i]] = i; 56 REP(i, 0, n) { 57 if (k) k--; 58 if (rnk[i] == 0) continue; 59 int j = sa[rnk[i] - 1]; 60 while (s[i + k] == s[j + k]) k++; 61 height[rnk[i]] = k; 62 } 63 } 64 65 bool judge(int key) { 66 int Min = 0x3f3f3f3f, Max = -0x3f3f3f3f; 67 REP(i, 1, n) { 68 if (height[i] >= key) { 69 Min = min(Min, min(sa[i - 1], sa[i])); 70 Max = max(Max, max(sa[i - 1], sa[i])); 71 } 72 else { 73 if (Max - Min > key) return true; 74 Min = 0x3f3f3f3f, Max = -0x3f3f3f3f; 75 } 76 } 77 if (Max - Min > key) return true; 78 else return false; 79 } 80 81 int a[maxn]; 82 int main() { 83 while (scanf("%d", &n) && n) { 84 REP(i, 0, n) scanf("%d", &a[i]); 85 if (n < 10) { puts("0"); continue; } 86 REP(i, 0, n - 1) s[i] = a[i + 1] - a[i] + 100; 87 s[n - 1] = 0; 88 build_sa(200), buildHeight(); 89 90 int low = 0, high = maxn - 1; 91 while (low < high - 1) { 92 int mid = low + high >> 1; 93 if (judge(mid)) low = mid; 94 else high = mid; 95 } 96 if (low + 1 < 5) puts("0"); 97 else printf("%d\n", low + 1); 98 } 99 return 0; 100 }
poj3261 <最大可重复公共子序列长度>
和上一题一样。不要忘记判judge()的最后一个！！！
1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 #define rep(i, a, b) for (int i = a; i <= b; i++) 5 #define drep(i, a, b) for (int i = a; i >= b; i--) 6 #define REP(i, a, b) for (int i = a; i < b; i++) 7 #define mp make_pair 8 #define pb push_back 9 #define clr(x) memset(x, 0, sizeof(x)) 10 #define xx first 11 #define yy second 12 13 using namespace std; 14 15 typedef long long i64; 16 typedef pair<int, int> pii; 17 const int inf = ~0U >> 1; 18 const i64 INF = ~0ULL >> 1; 19 //***************************** 20 21 const int maxn = 20005; 22 int hsh[maxn], cd; 23 int s[maxn], sa[maxn], t[maxn], t2[maxn], c[maxn], n; 24 void build_sa(int m) { 25 int *x = t, *y = t2; 26 REP(i, 0, m) c[i] = 0; 27 REP(i, 0, n) c[x[i] = s[i]]++; 28 REP(i, 1, m) c[i] += c[i - 1]; 29 drep(i, n - 1, 0) sa[--c[x[i]]] = i; 30 31 for (int k = 1; k <= n; k <<= 1) { 32 int p = 0; 33 REP(i, n - k, n) y[p++] = i; 34 REP(i, 0, n) if (sa[i] >= k) y[p++] = sa[i] - k; 35 36 REP(i, 0, m) c[i] = 0; 37 REP(i, 0, n) c[x[y[i]]]++; 38 REP(i, 1, m) c[i] += c[i - 1]; 39 drep(i, n - 1, 0) sa[--c[x[y[i]]]] = y[i]; 40 41 swap(x, y); 42 p = 1; x[sa[0]] = 0; 43 REP(i, 1, n) 44 x[sa[i]] = y[sa[i - 1]] == y[sa[i]] && y[sa[i - 1] + k] == y[sa[i] + k] ? p - 1 : p++; 45 if (p >= n) break; 46 m = p; 47 } 48 } 49 50 int rnk[maxn], height[maxn]; 51 void build_height() { 52 REP(i, 0, n) rnk[sa[i]] = i; 53 int k(0); 54 REP(i, 0, n - 1) { 55 if (k) k--; 56 int j = sa[rnk[i] - 1]; 57 while (s[i + k] == s[j + k]) k++; 58 height[rnk[i]] = k; 59 } 60 } 61 62 bool judge(int key, int k) { 63 int cnt(0); 64 REP(i, 2, n) { 65 if (height[i] >= key) cnt++; 66 else { 67 if (cnt + 1 >= k) return true; 68 cnt = 0; 69 } 70 } 71 if (cnt + 1 >= k) return true; 72 else return false; 73 } 74 75 int read() { 76 int l = 1, ret(0); 77 char ch = getchar(); 78 while (ch < '0' || ch > '9') {if (ch == '-') l = -1; ch = getchar(); } 79 while (ch >= '0' && ch <= '9') { ret = (ret << 1) + (ret << 3) + ch - '0'; ch = getchar(); } 80 return l * ret; 81 } 82 int a[maxn]; 83 int main() { 84 int k; 85 scanf("%d%d", &n, &k); 86 n++; 87 rep(i, 1, n) scanf("%d", &a[i]), hsh[cd++] = a[i]; 88 sort(hsh, hsh + n); cd = unique(hsh, hsh + cd) - hsh; 89 REP(i, 0, n) s[i] = lower_bound(hsh, hsh + cd, a[i + 1]) - hsh; 90 build_sa(cd); build_height(); 91 92 int low = 0, high = maxn; 93 while (low < high - 1) { 94 int mid = low + high >> 1; 95 if (judge(mid, k)) low = mid; 96 else high = mid; 97 } 98 printf("%d\n", low); 99 return 0; 100 }
spoj694 <不同子序列个数>
首先后缀数组，考虑按sa的顺序依次加入考虑，每加入一个，则以它开头的子串会增加len - sa[i]个，考虑这些被加入的字符串的重复情况，因为height往上递减，所以用它和以前的重复情况不如考虑它和它 - 1的重复情况，即减去height[i];
1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 #define rep(i, a, b) for (int i = a; i <= b; i++) 5 #define drep(i, a, b) for (int i = a; i >= b; i--) 6 #define REP(i, a, b) for (int i = a; i < b; i++) 7 #define mp make_pair 8 #define pb push_back 9 #define clr(x) memset(x, 0, sizeof(x)) 10 #define xx first 11 #define yy second 12 13 using namespace std; 14 15 typedef long long i64; 16 typedef pair<int, int> pii; 17 const int inf = ~0U >> 1; 18 const i64 INF = ~0ULL >> 1; 19 //***************************** 20 21 const int maxn = 1005; 22 char s[maxn]; 23 int sa[maxn], t[maxn], t2[maxn], c[maxn], n; 24 void build_sa(int m) { 25 int *x = t, *y = t2; 26 REP(i, 0, m) c[i] = 0; 27 REP(i, 0, n) c[x[i] = s[i]]++; 28 REP(i, 1, m) c[i] += c[i - 1]; 29 drep(i, n - 1, 0) sa[--c[x[i]]] = i; 30 31 for (int k = 1; k <= n; k <<= 1) { 32 int p = 0; 33 REP(i, n - k, n) y[p++] = i; 34 REP(i, 0, n) if (sa[i] >= k) y[p++] = sa[i] - k; 35 36 REP(i, 0, m) c[i] = 0; 37 REP(i, 0, n) c[x[y[i]]]++; 38 REP(i, 1, m) c[i] += c[i - 1]; 39 drep(i, n - 1, 0) sa[--c[x[y[i]]]] = y[i]; 40 41 swap(x, y); 42 p = 1; x[sa[0]] = 0; 43 REP(i, 1, n) 44 x[sa[i]] = y[sa[i]] == y[sa[i - 1]] && y[sa[i] + k] == y[sa[i - 1] + k] ? p - 1 : p++; 45 if (p >= n) break; 46 m = p; 47 } 48 } 49 50 int rnk[maxn], height[maxn]; 51 void build_height() { 52 int k(0); 53 REP(i, 0, n) rnk[sa[i]] = i; 54 REP(i, 0, n - 1) { 55 if (k) k--; 56 int j = sa[rnk[i] - 1]; 57 while (s[i + k] == s[j + k]) k++; 58 height[rnk[i]] = k; 59 } 60 } 61 62 int main() { 63 int T; 64 scanf("%d", &T); 65 while (T--) { 66 scanf("%s", s); 67 n = strlen(s) + 1; 68 build_sa('z' + 1); build_height(); 69 int ans(0); 70 REP(i, 1, n) { 71 ans += n - sa[i] - 1; 72 ans -= height[i]; 73 } 74 printf("%d\n", ans); 75 } 76 return 0; 77 }
2015年11月22日
ural1297 <最长回文子串>
将原字符串翻转后接在一开始的字符串后面，用rmq查询两个字符串的LCP，依次枚举每一个点即可。记住:n总比最后一个下标大2.
1 #include <bits/stdc++.h> 2 #define rep(i, a, b) for (int i = a; i <= b; i++) 3 #define drep(i, a, b) for (int i = a; i >= b; i--) 4 #define REP(i, a, b) for (int i = a; i < b; i++) 5 #define pb push_back 6 #define mp make_pair 7 #define clr(x) memset(x, 0, sizeof(x)) 8 #define xx first 9 #define yy second 10 using namespace std; 11 typedef long long i64; 12 typedef pair<int, int> pii; 13 const int inf = ~0U>>1; 14 const i64 INF = ~0ULL>>1; 15 //*************************** 16 17 const int maxn = 600005; 18 char s[maxn]; 19 int sa[maxn], t[maxn], t2[maxn], c[maxn], n; 20 void build_sa(int m) { 21 int *x = t, *y = t2; 22 REP(i, 0, m) c[i] = 0; 23 REP(i, 0, n) c[x[i] = s[i]]++; 24 REP(i, 1, m) c[i] += c[i - 1]; 25 drep(i, n - 1, 0) sa[--c[x[i]]] = i; 26 27 for (int k = 1; k <= n; k <<= 1) { 28 int p = 0; 29 REP(i, n - k, n) y[p++] = i; 30 REP(i, 0, n) if (sa[i] >= k) y[p++] = sa[i] - k; 31 32 REP(i, 0, m) c[i] = 0; 33 REP(i, 0, n) c[x[y[i]]]++; 34 REP(i, 1, m) c[i] += c[i - 1]; 35 drep(i, n - 1, 0) sa[--c[x[y[i]]]] = y[i]; 36 37 swap(x, y); 38 p = 1, x[sa[0]] = 0; 39 REP(i, 1, n) 40 x[sa[i]] = y[sa[i]] == y[sa[i - 1]] && y[sa[i] + k] == y[sa[i - 1] + k] ? p - 1 : p++; 41 if (p >= n) break; 42 m = p; 43 } 44 } 45 46 int rnk[maxn], height[maxn]; 47 void build_height() { 48 REP(i, 0, n) rnk[sa[i]] = i; 49 int k(0); 50 REP(i, 0, n - 1) { 51 if (k) k--; 52 int j = sa[rnk[i] - 1]; 53 while (s[i + k] == s[j + k]) k++; 54 height[rnk[i]] = k; 55 } 56 } 57 58 char str[maxn]; 59 60 int Min[maxn][32], Log[maxn]; 61 void build_st(int lim) { 62 Log[1] = 0; 63 rep(i, 2, lim) { 64 Log[i] = Log[i - 1]; 65 if (1 << Log[i] + 1 == i) Log[i]++; 66 } 67 drep(i, lim - 1, 1) { 68 Min[i][0] = height[i]; 69 for (int j = 1; i + (1 << j) - 1 < lim; j++) { 70 Min[i][j] = min(Min[i][j - 1], Min[i + (1 << j - 1)][j - 1]); 71 } 72 } 73 } 74 int query(int l, int r) { 75 int len = r - l + 1, k = Log[len]; 76 return min(Min[l][k], Min[r - (1 << k) + 1][k]); 77 } 78 79 int main() { 80 scanf("%s", str); 81 n = strlen(str); 82 memcpy(s, str, sizeof(str)); 83 s[n] = '$'; 84 rep(i, 1, n) s[n + i] = s[n - i]; 85 n++, n <<= 1; 86 build_sa('z' + 1); build_height(); 87 build_st(n); 88 int ans = -1, strt; 89 for (int i = 0; str[i]; i++) { 90 int j = n - 2 - i; 91 int tmp = query(min(rnk[i], rnk[j]) + 1, max(rnk[i], rnk[j])); 92 int len = (tmp << 1) - 1; 93 if (len > ans) ans = len, strt = i - tmp + 1; 94 j++; tmp = query(min(rnk[i], rnk[j]) + 1, max(rnk[i], rnk[j])); 95 len = tmp << 1; 96 if (len > ans) ans = len, strt = i - tmp; 97 } 98 while (ans--) putchar(s[strt++]); 99 return 0; 100 }
APIO2014 Palindromes <回文子串长度 * 这个子串出现次数的最大值> TLE
这个题应该Manacher+后缀数组，Manachar后面学，，我用后缀数组统计超时了。。。
1 #include <bits/stdc++.h> 2 #define rep(i, a, b) for (int i = a; i <= b; i++) 3 #define drep(i, a, b) for (int i = a; i >= b; i--) 4 #define REP(i, a, b) for (int i = a; i < b; i++) 5 #define pb push_back 6 #define mp make_pair 7 #define clr(x) memset(x, 0, sizeof(x)) 8 #define xx first 9 #define yy second 10 using namespace std; 11 typedef long long i64; 12 typedef pair<int, int> pii; 13 const int inf = ~0U>>1; 14 const i64 INF = ~0ULL>>1; 15 //*************************** 16 17 const int maxn = 600005; 18 char s[maxn]; 19 int sa[maxn], t[maxn], t2[maxn], c[maxn], n; 20 void build_sa(int m) { 21 int *x = t, *y = t2; 22 REP(i, 0, m) c[i] = 0; 23 REP(i, 0, n) c[x[i] = s[i]]++; 24 REP(i, 1, m) c[i] += c[i - 1]; 25 drep(i, n - 1, 0) sa[--c[x[i]]] = i; 26 27 for (int k = 1; k <= n; k <<= 1) { 28 int p = 0; 29 REP(i, n - k, n) y[p++] = i; 30 REP(i, 0, n) if (sa[i] >= k) y[p++] = sa[i] - k; 31 32 REP(i, 0, m) c[i] = 0; 33 REP(i, 0, n) c[x[y[i]]]++; 34 REP(i, 1, m) c[i] += c[i - 1]; 35 drep(i, n - 1, 0) sa[--c[x[y[i]]]] = y[i]; 36 37 swap(x, y); 38 p = 1, x[sa[0]] = 0; 39 REP(i, 1, n) 40 x[sa[i]] = y[sa[i]] == y[sa[i - 1]] && y[sa[i] + k] == y[sa[i - 1] + k] ? p - 1 : p++; 41 if (p >= n) break; 42 m = p; 43 } 44 } 45 46 int rnk[maxn], height[maxn]; 47 void build_height() { 48 REP(i, 0, n) rnk[sa[i]] = i; 49 int k(0); 50 REP(i, 0, n - 1) { 51 if (k) k--; 52 int j = sa[rnk[i] - 1]; 53 while (s[i + k] == s[j + k]) k++; 54 height[rnk[i]] = k; 55 } 56 } 57 58 char str[maxn]; 59 60 int Min[maxn][32], Log[maxn]; 61 void build_st(int lim) { 62 Log[1] = 0; 63 rep(i, 2, lim) { 64 Log[i] = Log[i - 1]; 65 if (1 << Log[i] + 1 == i) Log[i]++; 66 } 67 drep(i, lim - 1, 1) { 68 Min[i][0] = height[i]; 69 for (int j = 1; i + (1 << j) - 1 < lim; j++) { 70 Min[i][j] = min(Min[i][j - 1], Min[i + (1 << j - 1)][j - 1]); 71 } 72 } 73 } 74 int query(int l, int r) { 75 int len = r - l + 1, k = Log[len]; 76 return min(Min[l][k], Min[r - (1 << k) + 1][k]); 77 } 78 79 int search_above(int low, int high, int len, int pos) { 80 while (low < high - 1) { 81 int mid = low + high >> 1; 82 if (query(mid + 1, pos) >= len) high = mid; 83 else low = mid; 84 } 85 return high; 86 } 87 88 int search_behind(int low, int high, int len, int pos) { 89 while (low < high - 1) { 90 int mid = low + high >> 1; 91 if (query(pos + 1, mid) >= len) low = mid; 92 else high = mid; 93 } 94 return low; 95 } 96 97 int ans = -1; 98 void solve(int st, int len) { 99 while (1) { 100 int pos = rnk[st]; 101 int pos1 = search_above(0, pos, len, pos); 102 int pos2 = search_behind(pos, n, len, pos); 103 ans = max(ans, (pos2 - pos1 + 1) * len >> 1); 104 len -= 2; 105 st += 1; 106 if (len <= 0) break; 107 } 108 } 109 110 int main() { 111 scanf("%s", str); 112 n = strlen(str); 113 memcpy(s, str, sizeof(str)); 114 s[n] = '$'; 115 rep(i, 1, n) s[n + i] = s[n - i]; 116 n++, n <<= 1; 117 build_sa('z' + 1); build_height(); 118 build_st(n); 119 for (int i = 0; str[i]; i++) { 120 int j = n - 2 - i; 121 int tmp = query(min(rnk[i], rnk[j]) + 1, max(rnk[i], rnk[j])); 122 int len = (tmp << 1) - 1; 123 solve(i - tmp + 1, len); 124 j++; tmp = query(min(rnk[i], rnk[j]) + 1, max(rnk[i], rnk[j])); 125 len = tmp << 1; 126 solve(i - tmp, len); 127 } 128 printf("%d\n", ans); 129 return 0; 130 }
poj2406 <连续重复子串> TLE
倍增后缀数组TLE...原来是KMP做的。
暴力枚举长度k，然后查询rnk[n - k]和rnk[0]的LCP，若LCP >= n - k则当前k可行。只需预先处理出所有后缀和rnk[0]的最小值即可。
1 #include <bits/stdc++.h> 2 #define rep(i, a, b) for (int i = a; i <= b; i++) 3 #define drep(i, a, b) for (int i = a; i >= b; i--) 4 #define REP(i, a, b) for (int i = a; i < b; i++) 5 #define pb push_back 6 #define mp make_pair 7 #define clr(x) memset(x, 0, sizeof(x)) 8 #define xx first 9 #define yy second 10 using namespace std; 11 typedef long long i64; 12 typedef pair<int, int> pii; 13 const int inf = ~0U>>1; 14 const i64 INF = ~0ULL>>1; 15 //*************************** 16 17 const int maxn = 1000005; 18 char s[maxn]; 19 int sa[maxn], t[maxn], t2[maxn], c[maxn], n; 20 void build_sa(int m) { 21 int *x = t, *y = t2; 22 REP(i, 0, m) c[i] = 0; 23 REP(i, 0, n) c[x[i] = s[i]]++; 24 REP(i, 1, m) c[i] += c[i - 1]; 25 drep(i, n - 1, 0) sa[--c[x[i]]] = i; 26 27 for (int k = 1; k <= n; k++) { 28 int p(0); 29 REP(i, n - k, n) y[p++] = i; 30 REP(i, 0, n) if (sa[i] >= k) y[p++] = sa[i] - k; 31 32 REP(i, 0, m) c[i] = 0; 33 REP(i, 0, n) c[x[y[i]]]++; 34 REP(i, 1, m) c[i] += c[i - 1]; 35 drep(i, n - 1, 0) sa[--c[x[y[i]]]] = y[i]; 36 37 swap(x, y); 38 p = 1, x[sa[0]] = 0; 39 REP(i, 1, n) 40 x[sa[i]] = y[sa[i - 1]] == y[sa[i]] && y[sa[i - 1] + k] == y[sa[i] + k] ? p - 1 : p++; 41 if (p >= n) break; 42 m = p; 43 } 44 } 45 46 int rnk[maxn], height[maxn]; 47 void build_height() { 48 REP(i, 0, n) rnk[sa[i]] = i; 49 int k(0); 50 REP(i, 0, n - 1) { 51 if (k) k--; 52 int j = sa[rnk[i] - 1]; 53 while (s[i + k] == s[j + k]) k++; 54 height[rnk[i]] = k; 55 } 56 } 57 58 int Min[maxn]; 59 int main() { 60 while (1) { 61 scanf("%s", s); 62 if (s[0] == '.') break; 63 n = strlen(s) + 1; 64 build_sa('z' + 1), build_height(); 65 66 int pos = rnk[0]; 67 Min[pos - 1] = height[pos], Min[pos + 1] = height[pos + 1]; 68 drep(i, pos - 2, 0) Min[i] = min(Min[i + 1], height[i + 1]); 69 REP(i, pos + 2, n) Min[i] = min(Min[i - 1], height[i]); 70 71 int len = strlen(s); 72 rep(i, 1, len) { 73 if (len % i) continue; 74 if (Min[rnk[i]] >= len - i) { 75 printf("%d\n", len / i); 76 break; 77 } 78 } 79 } 80 return 0; 81 }
poj3693 <重复次数最多的连续重复子串>
先考虑重复两遍的情况，这个子串S中的[0], [L], [L * 2],[L * 3]...中至少有相邻两个被包含，考察 S[i * l] 和 S[(i + 1) * l] 的LCP，设其值为k，若 k%l有值，则可认为LCP多匹配了k%l，也可认为是少匹配了 l - k%l，向前枚举直到不符合.
1 #include <bits/stdc++.h> 2 #define rep(i, a, b) for (int i = a; i <= b; i++) 3 #define drep(i, a, b) for (int i = a; i >= b; i--) 4 #define REP(i, a, b) for (int i = a; i < b; i++) 5 #define pb push_back 6 #define mp make_pair 7 #define clr(x) memset(x, 0, sizeof(x)) 8 #define xx first 9 #define yy second 10 using namespace std; 11 typedef long long i64; 12 typedef pair<int, int> pii; 13 const int inf = ~0U>>1; 14 const i64 INF = ~0ULL>>1; 15 //*************************** 16 17 const int maxn = 200005; 18 char s[maxn]; 19 int sa[maxn], t[maxn], t2[maxn], c[maxn], n; 20 void build_sa(int m) { 21 int *x = t, *y = t2; 22 REP(i, 0, m) c[i] = 0; 23 REP(i, 0, n) c[x[i] = s[i]]++; 24 REP(i, 1, m) c[i] += c[i - 1]; 25 drep(i, n - 1, 0) sa[--c[x[i]]] = i; 26 27 for (int k = 1; k <= n; k <<= 1) { 28 int p(0); 29 REP(i, n - k, n) y[p++] = i; 30 REP(i, 0, n) if (sa[i] >= k) y[p++] = sa[i] - k; 31 32 REP(i, 0, m) c[i] = 0; 33 REP(i, 0, n) c[x[y[i]]]++; 34 REP(i, 1, m) c[i] += c[i - 1]; 35 drep(i, n - 1, 0) sa[--c[x[y[i]]]] = y[i]; 36 37 swap(x, y); 38 p = 1, x[sa[0]] = 0; 39 REP(i, 1, n) 40 x[sa[i]] = y[sa[i]] == y[sa[i - 1]] && y[sa[i] + k] == y[sa[i - 1] + k] ? p - 1 : p++; 41 if (p >= n) break; 42 m = p; 43 } 44 } 45 46 int rnk[maxn], height[maxn]; 47 void build_height() { 48 REP(i, 0, n) rnk[sa[i]] = i; 49 int k(0); 50 REP(i, 0, n - 1) { 51 if (k) k--; 52 int j = sa[rnk[i] - 1]; 53 while (s[i + k] == s[j + k]) k++; 54 height[rnk[i]] = k; 55 } 56 } 57 58 int Min[maxn][32], Log[maxn]; 59 void build_st(int lim) { 60 Log[1] = 0; 61 rep(i, 2, lim) { 62 Log[i] = Log[i - 1]; 63 if (1 << Log[i] + 1 == i) Log[i]++; 64 } 65 drep(i, lim, 1) { 66 Min[i][0] = height[i]; 67 for (int j = 1; i + (1 << j) - 1 <= lim; j++) 68 Min[i][j] = min(Min[i][j - 1], Min[i + (1 << j - 1)][j - 1]); 69 } 70 } 71 int query(int l, int r) { 72 int len = r - l + 1, k = Log[len]; 73 return min(Min[l][k], Min[r - (1 << k) + 1][k]); 74 } 75 76 int main() { 77 int T(0); 78 while(~scanf("%s",s) && s[0]!='#') { 79 T++; 80 printf("Case %d: ", T); 81 n = strlen(s) + 1; 82 build_sa('z' + 1), build_height(); 83 build_st(n - 1); 84 int len = strlen(s); 85 if (len == 1) { puts(s); return 0; } 86 int ans = -1, st = inf, pril = 0; 87 rep(l, 1, len / 2) { 88 for (int i = 0; (i + 1) * l < len; i++) { 89 int q1 = i * l, q2 = (i + 1) * l; 90 int cur_ans = query(min(rnk[q1], rnk[q2]) + 1, max(rnk[q1], rnk[q2])); 91 int cur_st = q1; 92 int r = l - cur_ans % l; 93 cur_ans = cur_ans / l + 1; 94 int cnt(0); 95 for (int j = q1 - 1; j > q1 - l && s[j] == s[j + l] && j >= 0; j--) { 96 cnt++; 97 if (cnt == r) { 98 cur_ans++; 99 cur_st = j; 100 } 101 if (rnk[j] < rnk[cur_st]) cur_st = j; 102 } 103 if (cur_ans > ans || cur_ans == ans && rnk[cur_st] < rnk[st]) { 104 ans = cur_ans; 105 st = cur_st; 106 pril = l; 107 } 108 } 109 } 110 int pri = ans * pril; 111 for (int i = st, j = 0; j < pri; j++) printf("%c", s[i + j]); 112 puts(""); 113 } 114 return 0; 115 }
2015年11月23日
spoj687 <重复次数最多的连续子串长度>
和上一道题一样，只是不用一位一位往前移动了，少了 l - (k % l) 的值，所以在枚举相邻的i时，也向前l-k%l位，查询LCP即可。
1 #include <bits/stdc++.h> 2 #define rep(i, a, b) for (register int i = a; i <= b; i++) 3 #define drep(i, a, b) for (int i = a; i >= b; i--) 4 #define REP(i, a, b) for (int i = a; i < b; i++) 5 #define pb push_back 6 #define mp make_pair 7 #define clr(x) memset(x, 0, sizeof(x)) 8 #define xx first 9 #define yy second 10 using namespace std; 11 typedef long long i64; 12 typedef pair<int, int> pii; 13 const int inf = 0x3f3f3f3f; 14 const i64 INF = ~0ULL>>1; 15 //*************************** 16 17 const int maxn = 50005; 18 char s[maxn]; 19 int sa[maxn], t[maxn], t2[maxn], c[maxn], n; 20 void build_sa(int m) { 21 int *x = t, *y = t2; 22 REP(i, 0, m) c[i] = 0; 23 REP(i, 0, n) c[x[i] = s[i]]++; 24 REP(i, 1, m) c[i] += c[i - 1]; 25 drep(i, n - 1, 0) sa[--c[x[i]]] = i; 26 27 for (int k = 1; k <= n; k <<= 1) { 28 int p(0); 29 REP(i, n - k, n) y[p++] = i; 30 REP(i, 0, n) if (sa[i] >= k) y[p++] = sa[i] - k; 31 32 REP(i, 0, m) c[i] = 0; 33 REP(i, 0, n) c[x[y[i]]]++; 34 REP(i, 1, m) c[i] += c[i - 1]; 35 drep(i, n - 1, 0) sa[--c[x[y[i]]]] = y[i]; 36 37 swap(x, y); 38 p = 1, x[sa[0]] = 0; 39 REP(i, 1, n) 40 x[sa[i]] = y[sa[i]] == y[sa[i - 1]] && y[sa[i] + k] == y[sa[i - 1] + k] ? p - 1 : p++; 41 if (p >= n) break; 42 m = p; 43 } 44 } 45 46 int rnk[maxn], height[maxn]; 47 void build_height() { 48 REP(i, 0, n) rnk[sa[i]] = i; 49 int k(0); 50 REP(i, 0, n - 1) { 51 if (k) k--; 52 int j = sa[rnk[i] - 1]; 53 while (s[i + k] == s[j + k]) k++; 54 height[rnk[i]] = k; 55 } 56 } 57 58 int Min[maxn][32], Log[maxn]; 59 void build_st(int lim) { 60 Log[1] = 0; 61 rep(i, 2, lim) { 62 Log[i] = Log[i - 1]; 63 if (1 << Log[i] + 1 == i) Log[i]++; 64 } 65 drep(i, lim, 1) { 66 Min[i][0] = height[i]; 67 for (int j = 1; i + (1 << j) - 1 <= lim; j++) 68 Min[i][j] = min(Min[i][j - 1], Min[i + (1 << j - 1)][j - 1]); 69 } 70 } 71 int query(int l, int r) { 72 int len = r - l + 1, k = Log[len]; 73 return min(Min[l][k], Min[r - (1 << k) + 1][k]); 74 } 75 76 char ch[5]; 77 int main() { 78 int T; 79 scanf("%d", &T); 80 while (T--) { 81 scanf("%d", &n); 82 REP(i, 0, n) { scanf("%s", ch); s[i] = ch[0]; } 83 n++; 84 build_sa('z' + 1), build_height(); 85 build_st(n - 1); 86 int len = n - 1; 87 int ans = -1; 88 rep(l, 1, len >> 1) { 89 for (int i = 0; (i + 1) * l < len; i++) { 90 int LCP = query(min(rnk[i * l], rnk[(i + 1) * l]) + 1, max(rnk[i * l], rnk[(i + 1) * l])); 91 ans = max(ans, LCP / l + 1); 92 int q1 = i * l - (l - LCP % l); 93 if (q1 < 0) continue; 94 LCP = query(min(rnk[q1], rnk[q1 + l]) + 1, max(rnk[q1], rnk[q1 + l])); 95 ans = max(ans, LCP / l + 1); 96 } 97 } 98 if (ans == -1) ans = 1; 99 printf("%d\n", ans); 100 } 101 }
2015年11月24日
poj2774 <两个字符串的最长公共子串长度>
首先把两个字符串拼在一起，观察height数组，sa数组中排名相邻的两个字符串若不同属于一个字符串，那么这个height被纳入考虑。
不相邻的不同属于一个字符串的不需要考虑，假设字符串a排名靠前，字符串b排名靠后，考虑b串，加入b串上面和它不同属于一个串，那么LCP(a, b) 没有LCP(b - 1, b)优；假如b - 1这个串和b同属于一个串，那么LCP(a, b) 没有 LCP(a, b - 1)优，所以只需要考虑相邻的即可。
1 #include <bits/stdc++.h> 2 #define rep(i, a, b) for (register int i = a; i <= b; i++) 3 #define drep(i, a, b) for (int i = a; i >= b; i--) 4 #define REP(i, a, b) for (int i = a; i < b; i++) 5 #define pb push_back 6 #define mp make_pair 7 #define clr(x) memset(x, 0, sizeof(x)) 8 #define xx first 9 #define yy second 10 using namespace std; 11 typedef long long i64; 12 typedef pair<int, int> pii; 13 const int inf = 0x3f3f3f3f; 14 const i64 INF = ~0ULL>>1; 15 //*************************** 16 17 const int maxn = 200005; 18 char s[maxn]; 19 int sa[maxn], t[maxn], t2[maxn], c[maxn], n; 20 void build_sa(int m) { 21 int *x = t, *y = t2; 22 REP(i, 0, m) c[i] = 0; 23 REP(i, 0, n) c[x[i] = s[i]]++; 24 REP(i, 1, m) c[i] += c[i - 1]; 25 drep(i, n - 1, 0) sa[--c[x[i]]] = i; 26 27 for (int k = 1; k <= n; k <<= 1) { 28 int p(0); 29 REP(i, n - k, n) y[p++] = i; 30 REP(i, 0, n) if (sa[i] >= k) y[p++] = sa[i] - k; 31 32 REP(i, 0, m) c[i] = 0; 33 REP(i, 0, n) c[x[y[i]]]++; 34 REP(i, 1, m) c[i] += c[i - 1]; 35 drep(i, n - 1, 0) sa[--c[x[y[i]]]] = y[i]; 36 37 swap(x, y); 38 p = 1; x[sa[0]] = 0; 39 REP(i, 1, n) 40 x[sa[i]] = y[sa[i]] == y[sa[i - 1]] && y[sa[i] + k] == y[sa[i - 1] + k] ? p - 1 : p++; 41 if (p >= n) break; 42 m = p; 43 } 44 } 45 46 int rnk[maxn], height[maxn]; 47 void build_height() { 48 REP(i, 0, n) rnk[sa[i]] = i; 49 int k(0); 50 REP(i, 0, n - 1) { 51 if (k) k--; 52 int j = sa[rnk[i] - 1]; 53 while (s[i + k] == s[j + k]) k++; 54 height[rnk[i]] = k; 55 } 56 } 57 58 char str[maxn >> 1]; 59 int main() { 60 scanf("%s", str); 61 memcpy(s, str, sizeof(str)); 62 int cur = strlen(str); 63 int k = cur; 64 scanf("%s", str); 65 s[cur++] = '$'; 66 for (int i = 0; str[i]; i++) s[cur++] = str[i]; 67 n = cur + 1; 68 build_sa('z' + 1); build_height(); 69 int ans = -1; 70 REP(i, 1, n) { 71 int l = sa[i - 1], r = sa[i]; 72 if (l > r) swap(l, r); 73 if (l < k && r > k) ans = max(ans, height[i]); 74 } 75 printf("%d\n", ans); 76 return 0; 77 }
ural1517 <两个字符串最长公共子串长度>
和上一道题一样，不再赘述。
1 #include <bits/stdc++.h> 2 #define rep(i, a, b) for (register int i = a; i <= b; i++) 3 #define drep(i, a, b) for (int i = a; i >= b; i--) 4 #define REP(i, a, b) for (int i = a; i < b; i++) 5 #define pb push_back 6 #define mp make_pair 7 #define clr(x) memset(x, 0, sizeof(x)) 8 #define xx first 9 #define yy second 10 using namespace std; 11 typedef long long i64; 12 typedef pair<int, int> pii; 13 const int inf = 0x3f3f3f3f; 14 const i64 INF = ~0ULL>>1; 15 //*************************** 16 17 const int maxn = 200005; 18 char s[maxn]; 19 int sa[maxn], t[maxn], t2[maxn], c[maxn], n; 20 void build_sa(int m) { 21 int *x = t, *y = t2; 22 REP(i, 0, m) c[i] = 0; 23 REP(i, 0, n) c[x[i] = s[i]]++; 24 REP(i, 1, m) c[i] += c[i - 1]; 25 drep(i, n - 1, 0) sa[--c[x[i]]] = i; 26 27 for (int k = 1; k <= n; k <<= 1) { 28 int p(0); 29 REP(i, n - k, n) y[p++] = i; 30 REP(i, 0, n) if (sa[i] >= k) y[p++] = sa[i] - k; 31 32 REP(i, 0, m) c[i] = 0; 33 REP(i, 0, n) c[x[y[i]]]++; 34 REP(i, 1, m) c[i] += c[i - 1]; 35 drep(i, n - 1, 0) sa[--c[x[y[i]]]] = y[i]; 36 37 swap(x, y); 38 p = 1; x[sa[0]] = 0; 39 REP(i, 1, n) 40 x[sa[i]] = y[sa[i]] == y[sa[i - 1]] && y[sa[i] + k] == y[sa[i - 1] + k] ? p - 1 : p++; 41 if (p >= n) break; 42 m = p; 43 } 44 } 45 46 int rnk[maxn], height[maxn]; 47 void build_height() { 48 REP(i, 0, n) rnk[sa[i]] = i; 49 int k(0); 50 REP(i, 0, n - 1) { 51 if (k) k--; 52 int j = sa[rnk[i] - 1]; 53 while (s[i + k] == s[j + k]) k++; 54 height[rnk[i]] = k; 55 } 56 } 57 58 int main() { 59 int len; 60 scanf("%d", &len); 61 scanf("%s", s); 62 s[len] = '$'; 63 scanf("%s", s + len + 1); 64 n = (len << 1) + 2; 65 build_sa('z' + 1), build_height(); 66 int ans = -1, st; 67 REP(i, 1, n) { 68 int q1 = sa[i], q2 = sa[i - 1]; 69 if (q1 > q2) swap(q1, q2); 70 if (q1 < len && q2 > len) { 71 if (height[i] > ans) ans = height[i], st = q1; 72 } 73 } 74 while (ans--) putchar(s[st++]); 75 return 0; 76 }
