poj2774:对后缀数组的复仇，O(nlog n)做法
话说上次发了一篇文章..A掉了2774这道水题，但是！用的是O(nlog^2 n)做法，有些不服..
于是本人又研究了一次后缀数组的O(nlog n)做法，终于在昨晚领悟了！特A一题..终于可以教会师弟师妹们这一种数据结构了..哭
直接上代码吧..
1 /* 2 Source Code 3 4 Problem: 2774 User: aclolicon 5 Memory: 5212K Time: 610MS 6 Language: G++ Result: Accepted 7 Source Code*/ 8 #include<cstdio> 9 #include<iostream> 10 #include<cstring> 11 #define MAXN 200500 12 using namespace std; 13 char s[MAXN], s2[MAXN]; 14 int n, len; 15 int sa[MAXN], c[MAXN], t[MAXN], t2[MAXN], rank[MAXN], height[MAXN]; 16 void buildsa(){ 17 int *x = t, *y = t2, m = 300; 18 memset(c, 0, sizeof(c)); 19 20 for (int i = 0; i < n; i++) c[x[i] = s[i]]++; 21 for (int i = 1; i < m; i++) c[i] += c[i - 1]; 22 for (int i = n - 1; i >= 0; i--) sa[--c[x[i]]] = i; 23 for (int k = 1; k <= n; k *= 2){ 24 int p = 0; 25 for (int i = n - k; i < n; i++) y[p++] = i; 26 for (int i = 0; i < n; i++) if (sa[i] >= k) y[p++] = sa[i] - k; 27 memset(c, 0, sizeof(c)); 28 for (int i = 0; i < n; i++) c[x[y[i]]]++; 29 for (int i = 1; i < m; i++) c[i] += c[i - 1]; 30 for (int i = n - 1; i >= 0; i--) sa[--c[x[y[i]]]] =y[i]; 31 swap(x, y); 32 p = 1; 33 x[sa[0]] = 0; 34 for (int i = 1; i < n; i++) x[sa[i]] = (y[sa[i - 1]] == y[sa[i]] && y[sa[i - 1] + k] == y[sa[i] + k])? p - 1: p++; 35 if (p >= n) break; 36 m = p; 37 } 38 } 39 40 void geth(){ 41 n = strlen(s); 42 for (int i = 0; i <= n; i++) rank[sa[i]] = i; 43 int h = 0; 44 height[0] = 0; 45 for (int i = 0; i < n; i++){ 46 int j = sa[rank[i] - 1]; 47 if (h > 0) h--; 48 for (; j + h < n && i + h < n; h++){ 49 if (s[j + h] != s[i + h]) break; 50 } 51 height[rank[i] - 1] = h; 52 } 53 } 54 int main(){ 55 while(scanf("%s", s) > 0){ 56 scanf("%s", s2) > 0; 57 len = strlen(s2); 58 n = strlen(s); 59 if (s2 > s) swap(s, s2); 60 s[len] = '$'; 61 for (int i = len + 1; i < len + n + 1; i++) s[i] = s2[i - len - 1]; 62 n = strlen(s) + 1; 63 buildsa(); 64 geth(); 65 int maxh = -1; 66 for (int i = 0; i < n; i++){ 67 if((sa[i] < len) != (sa[i + 1] < len)){ 68 maxh = max(maxh, height[i]); 69 } 70 } 71 printf("%d", maxh); 72 } 73 }
