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POJ 2406 Power String 后缀数组
这题曾经用KMP做过,用KMP 做非常的简单,h函数自带的找循环节功能。
用后缀数组的话,首先枚举循环节长度k,然后比较LCP(suffix(k + 1), suffix(0)) 是否等于len - k, 如果相等显然k就是一个循环节。
得到LCP的话可以通过预处理出所有点和0的lcp就好了。另外倍增法构造后缀数组还有用RMQ来搞lcp nlogn是不行的,会超时,所以可以dc3走起了。。
#include <cstdio>#include <cstring>#include <algorithm>#include <queue>#include <stack>#include <map>#include <set>#include <climits>#include <string>using namespace std; #define MP make_pair#define PB push_backtypedef long long LL;typedef unsigned long long ULL;typedef vector<int> VI;typedef pair<int, int> PII;typedef pair<double, double> PDD;const int INF = INT_MAX / 3;const double eps = 1e-8;const LL LINF = 1e17;const double DINF = 1e60;const int maxn = 4e6 + 50;//以下是DC3算法求后缀数组#define F(x) ((x) / 3 + ((x) % 3 == 1 ? 0 : tb))#define G(x) ((x) < tb ? (x) * 3 + 1 : ((x) - tb) * 3 + 2)int wa[maxn], wb[maxn], wv[maxn], ws[maxn];int c0(int *r, int a, int b) { return r[a] == r[b] && r[a + 1] == r[b + 1] && r[a + 2] == r[b + 2];}int c12(int k, int *r, int a, int b) { if (k == 2) return r[a] < r[b] || r[a] == r[b] && c12(1, r, a + 1, b + 1); else return r[a] < r[b] || r[a] == r[b] && wv[a + 1] < wv[b + 1];}void sort(int *r, int *a, int *b, int n, int m) { int i; for (i = 0; i < n; i++) wv[i] = r[a[i]]; for (i = 0; i < m; i++) ws[i] = 0; for (i = 0; i < n; i++) ws[wv[i]]++; for (i = 1; i < m; i++) ws[i] += ws[i - 1]; for (i = n - 1; i >= 0; i--) b[--ws[wv[i]]] = a[i];}void dc3(int *r, int *sa, int n, int m) { int i, j, *rn = r + n, *san = sa + n, ta = 0, tb = (n + 1) / 3, tbc = 0, p; r[n] = r[n + 1] = 0; for (i = 0; i < n; i++) if (i % 3 != 0) wa[tbc++] = i; sort(r + 2, wa, wb, tbc, m); sort(r + 1, wb, wa, tbc, m); sort(r, wa, wb, tbc, m); for (p = 1, rn[F(wb[0])] = 0, i = 1; i < tbc; i++) rn[F(wb[i])] = c0(r, wb[i - 1], wb[i]) ? p - 1 : p++; if (p < tbc) dc3(rn, san, tbc, p); else for (i = 0; i < tbc; i++) san[rn[i]] = i; for (i = 0; i < tbc; i++) if (san[i] < tb) wb[ta++] = san[i] * 3; if (n % 3 == 1) wb[ta++] = n - 1; sort(r, wb, wa, ta, m); for (i = 0; i < tbc; i++) wv[wb[i] = G(san[i])] = i; for (i = 0, j = 0, p = 0; i < ta && j < tbc; p++) sa[p] = c12(wb[j] % 3, r, wa[i], wb[j]) ? wa[i++] : wb[j++]; for (; i < ta; p++) sa[p] = wa[i++]; for (; j < tbc; p++) sa[p] = wb[j++];}int Rank[maxn], height[maxn];void calheight(int *r, int *sa, int n) { int i, j, k = 0; for (i = 1; i <= n; i++) Rank[sa[i]] = i; for (i = 0; i < n; height[Rank[i++]] = k) for (k ? k-- : 0, j = sa[Rank[i] - 1]; r[i + k] == r[j + k]; k++);}#undef F#undef Gchar buf[maxn];int str[maxn], sa[maxn], len;int main() { while(scanf("%s", buf), buf[0] != ‘.‘) { len = strlen(buf); for(int i = 0; i < len; i++) str[i] = buf[i] + 1; str[len] = 0; dc3(str, sa, len + 1, 200); calheight(str, sa, len); for(int i = Rank[0] - 1, val = height[Rank[0]]; i > 0; i--) { int tmp = val; val = min(val, height[i]), height[i] = tmp; } for(int i = Rank[0] + 1; i + 1 < len; i++) height[i + 1] = min(height[i + 1], height[i]); int ans = 1; for(int i = 0; i < len; i++) if(len % (i + 1) == 0) { if(height[Rank[i + 1]] == len - i - 1) { ans = len / (i + 1); break; } } printf("%d\n", ans); } return 0;}
POJ 2406 Power String 后缀数组
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