POJ2774 后缀数组

 1 /*
 2 后缀数组就是套模板求先应得数组，这题用到了两个数组，分别是sa[],height[];
 3 sa[i]表示所有后缀按字典数排序后以s[i]开始的后缀排在第i位。height[i]表示
 4 字典数为i和i-1后缀的的最长串的前缀。
 5 */
 6 
 7 #include <iostream>
 8 #include <stdio.h>
 9 #include <string.h>
10 #include <algorithm>
11 using namespace std;
12 const int MAX = 2e5+10;
13 char str[MAX];
14 int s[MAX], sa[MAX], t[MAX], t2[MAX], c[MAX], n;
15 int rank1[MAX], height[MAX];
16 
17 void build_sa(int m) {
18     int *x = t, *y = t2;
19     for(int i = 0; i < m; i ++) c[i] = 0;
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 <<= 1) {
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         for(int i = 0; i < m; i ++) c[i] = 0;
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 ++)
35             x[sa[i]] = y[sa[i-1]] == y[sa[i]] && y[sa[i-1]+k] == y[sa[i]+k]?p-1:p++;
36         if(p >= n)break;
37         m = p;
38     }
39 }
40 void getHeight() {
41     int k = 0;
42     for(int i = 0; i < n; i ++) rank1[sa[i]] = i;
43     for(int i = 0; i < n; i ++) {
44         if(k) k--;
45         int j = sa[rank1[i]-1];
46         while(s[i+k] == s[j+k]) k++;
47         height[rank1[i]] = k;
48     }
49 }
50 int main() {
51     n = 0;
52     int l1, l2;
53     scanf("%s",str);
54     l1 = strlen(str);
55     for(int i = 0; i < l1; i ++) s[n++] = str[i] - ‘a‘ + 1;
56     s[n++] = 28;
57     scanf("%s",str);
58     l2 = strlen(str);
59     for(int i = 0; i < l2; i ++) s[n++] = str[i] - ‘a‘ + 1;
60     s[n++] = 0;
61     build_sa(30);
62     getHeight();
63     int maxn = 0;
64     for(int i = 2; i < n; i ++) {
65         if(maxn < height[i]) {
66             if(sa[i] >= 0 && sa[i] < l1 && sa[i-1] > l1)
67                 maxn = height[i];
68             if(sa[i-1] >= 0 && sa[i-1] < l1 && sa[i] > l1)
69                 maxn = height[i];
70         }
71     }
72     printf("%d\n",maxn);
73     return 0;
74 }