倍增算法（da）
1 #define maxn 1000001
2 int wa[maxn],wb[maxn],wv[maxn],wss[maxn];
3
4 int cmp(int *r, int a, int b, int l) {
5 return r[a] == r[b] && r[a+l] == r[b+l];
6 }
7
8 void da(int *r, int *sa, int n, int m) {
9 int i, j, p, *x = wa, *y = wb, *t;
10 for(i = 0; i < m; i++) wss[i] = 0;
11 for(i = 0; i < n; i++) wss[x[i]=r[i]]++;
12 for(i = 1; i < m; i++) wss[i] += wss[i-1];
13 for(i = n-1; i >= 0; i--) sa[--wss[x[i]]] = i;
14 for(j = 1, p = 1; p < n; j *= 2, m = p) {
15 for(p = 0, i = n-j; i < n; i++) y[p++] = i;
16 for(i = 0; i < n; i++) if(sa[i] >= j) y[p++] = sa[i] - j;
17 for(i = 0; i < n; i++) wv[i] = x[y[i]];
18 for(i = 0; i < m; i++) wss[i] = 0;
19 for(i = 0; i < n; i++) wss[wv[i]] ++;
20 for(i = 1; i < m; i++) wss[i] += wss[i-1];
21 for(i = n-1; i>=0; i--) sa[--wss[wv[i]]] = y[i];
22 for(t = x, x = y, y = t, p = 1, x[sa[0]] = 0, i = 1; i < n; i++)
23 x[sa[i]]=cmp(y,sa[i-1],sa[i],j)?p-1:p++;
24 }
25 return;
26 }
27
28 int rank[maxn], height[maxn];
29 void calheight(int *r, int *sa, int n) {
30 int i, j, k = 0;
31 for(i = 1; i <= n; i++) rank[sa[i]] = i;
32 for(i = 0; i < n; height[rank[i++]] = k)
33 for(k ? k-- : 0, j = sa[rank[i]-1]; r[i+k] == r[j+k]; k++);
34 return;
35 }
36
37 int RMQ[maxn];
38 int mm[maxn];
39 int best[20][maxn];
40 void initRMQ(int n) {
41 int i, j, a, b;
42 for(mm[0] =- 1, i = 1; i <= n; i++)
43 mm[i] = ((i&(i-1)) == 0) ? mm[i-1]+1 : mm[i-1];
44 for(i = 1; i <= n; i++) best[0][i] = i;
45 for(i = 1; i <= mm[n]; i++)
46 for(j = 1; j <= n+1-(1<<i); j++) {
47 a = best[i-1][j];
48 b = best[i-1][j+(1<<(i-1))];
49 if(RMQ[a] < RMQ[b]) best[i][j] = a;
50 else best[i][j] = b;
51 }
52 return;
53 }
54
55 int askRMQ(int a, int b) {
56 int t;
57 t = mm[b-a+1];
58 b -= (1<<t)-1;
59 a = best[t][a];
60 b = best[t][b];
61 return RMQ[a] < RMQ[b] ? a : b;
62 }
63 int lcp(int a, int b) {
64 int t;
65 a = rank[a];
66 b = rank[b];
67 if(a > b) t = a, a = b, b = t;
68 return(height[askRMQ(a+1, b)]);
69 }
70
71 int sa[maxn], r[maxn];
DC3
1 #define maxn 1000009
2 #define F(x) ((x)/3+((x)%3 == 1?0:tb))
3 #define G(x) ((x)<tb ? (x)*3+1 : ((x)-tb)*3+2)
4
5 int wa[maxn], wb[maxn], wv[maxn], wss[maxn];
6 int c0(int *r, int a, int b) {
7 return r[a] == r[b] && r[a+1] == r[b+1] && r[a+2] == r[b+2];
8 }
9 int c12(int k, int *r, int a, int b) {
10 if(k == 2) return r[a] < r[b] || r[a] == r[b] && c12(1, r, a+1, b+1);
11 else return r[a] < r[b] || r[a] == r[b] && wv[a+1] < wv[b+1];
12 }
13
14 void sort(int *r, int *a, int *b, int n, int m) {
15 int i;
16 for(i = 0; i < n; i++) wv[i] = r[a[i]];
17 for(i = 0; i < m; i++) wss[i] = 0;
18 for(i = 0; i < n; i++) wss[wv[i]] ++;
19 for(i = 1; i < m; i++) wss[i] += wss[i-1];
20 for(i = n-1; i >= 0; i--) b[--wss[wv[i]]] = a[i];
21 return;
22 }
23
24 void dc3(int *r, int *sa, int n, int m) {
25 int i, j, *rn = r+n, *san = sa+n, ta = 0, tb = (n+1)/3, tbc=0, p;
26 r[n] = r[n+1] = 0;
27 for(i = 0; i < n; i++) if(i%3 != 0) wa[tbc++] = i;
28 sort(r+2, wa, wb, tbc, m);
29 sort(r+1, wb, wa, tbc, m);
30 sort(r, wa, wb, tbc, m);
31 for(p = 1, rn[F(wb[0])] = 0, i = 1; i < tbc; i++)
32 rn[F(wb[i])] = c0(r, wb[i-1], wb[i]) ? p-1 : p++;
33 if(p < tbc) dc3(rn, san, tbc, p);
34 else for(i = 0; i < tbc; i++) san[rn[i]] = i;
35 for(i = 0; i < tbc; i++) if(san[i] < tb) wb[ta++] = san[i]*3;
36 if(n%3 == 1) wb[ta++] = n-1;
37 sort(r, wb, wa, ta, m);
38 for(i = 0; i < tbc; i++) wv[wb[i]=G(san[i])] = i;
39 for(i = 0, j = 0, p = 0; i < ta && j < tbc; p++)
40 sa[p] = c12(wb[j]%3, r, wa[i], wb[j]) ? wa[i++] : wb[j++];
41 for(; i < ta; p++) sa[p] = wa[i++];
42 for(; j < tbc; p++) sa[p] = wb[j++];
43 return;
44 }
45
46 int rank[maxn*3], height[maxn*3];
47 void calheight(int *r, int *sa, int n) {
48 int i, j, k = 0;
49 for(i = 1; i <= n; i++) rank[sa[i]] = i;
50 for(i = 0; i < n; height[rank[i++]] = k)
51 for(k ? k-- : 0, j = sa[rank[i]-1]; r[i+k] == r[j+k]; k++);
52 return;
53 }
54
55 int RMQ[maxn];
56 int mm[maxn];
57 int best[20][maxn];
58 void initRMQ(int n) {
59 int i, j, a, b;
60 for(mm[0] =- 1, i = 1; i <= n; i++)
61 mm[i] = ((i & (i-1)) == 0) ? mm[i-1]+1 : mm[i-1];
62 for(i = 1; i <= n; i++) best[0][i] = i;
63 for(i = 1; i <= mm[n]; i++)
64 for(j = 1; j <= n+1-(1<<i); j++) {
65 a = best[i-1][j];
66 b = best[i-1][j+(1<<(i-1))];
67 if(RMQ[a] < RMQ[b]) best[i][j] = a;
68 else best[i][j] = b;
69 }
70 return;
71 }
72
73 int askRMQ(int a, int b) {
74 int t;
75 t = mm[b-a+1];
76 b -= (1<<t)-1;
77 a = best[t][a];
78 b = best[t][b];
79 return RMQ[a] < RMQ[b] ? a : b;
80 }
81
82 int lcp(int a, int b) {
83 int t;
84 a = rank[a];
85 b = rank[b];
86 if(a > b) t = a, a = b, b = t;
87 return(height[askRMQ(a+1, b)]);
88 }
89
90 int sa[maxn*3], r[maxn];
-------------------------------------------------------
http://www.spoj.com/problems/DISUBSTR/
spoj 694
求不同子串的个数
1 #include <iostream>
2 #include <string.h>
3 using namespace std;
4
5 const int maxn = 1009;
6 int a[maxn], sa[maxn], len;
7 char str[maxn];
8
9 int wa[maxn], wb[maxn], wv[maxn], wws[maxn];
10 int cmp(int *r, int a, int b, int l){
11 return r[a] == r[b] && r[a+l] == r[b+l];
12 }
13 //sa from 0 -> n-1
14 void da(int *r, int *sa, int n, int m){
15 int i, j, p, *x = wa, *y = wb, *t;
16 for(i = 0; i < m; i++) wws[i] = 0;
17 for(i = 0; i < n; i++) wws[x[i] = r[i]]++;
18 for(i = 1; i < m; i++) wws[i] += wws[i-1];
19 for(i = n-1; i >= 0; i--) sa[--wws[x[i]]] = i;
20 for(j = 1,p = 1; p < n; j *= 2, m = p){
21 for(p = 0,i = n-j; i<n;i++) y[p++] = i;
22 for(i = 0; i < n; i++) if(sa[i] >= j) y[p++] = sa[i]-j;
23 for(i = 0; i < n; i++) wv[i] = x[y[i]];
24 for(i = 0; i < m; i++) wws[i] = 0;
25 for(i = 0; i < n; i++) wws[wv[i]] ++;
26 for(i = 1; i < m; i++) wws[i] += wws[i-1];
27 for(i = n-1; i >= 0; i--) sa[--wws[wv[i]]] = y[i];
28 for(t = x, x = y, y = t, p = 1, x[sa[0]] = 0, i = 1; i < n; i++)
29 x[sa[i]] = cmp(y,sa[i-1], sa[i],j) ? p-1 : p++;
30 }
31 return ;
32 }
33
34 int Rank[maxn], height[maxn];
35 void calheight(int *r, int *sa, int n){
36 int i, j, k = 0;
37 for(i = 1; i <= n; i++) Rank[sa[i]] = i;
38 for(i = 0; i < n; height[Rank[i++]] = k)
39 for(k ? k-- : 0, j = sa[Rank[i]-1]; r[i+k] == r[j+k]; k++);
40 return;
41 }
42
43
44 int main(){
45 int t;
46 scanf("%d", &t);
47 while(t--){
48 scanf("%s", str);
49 int len = strlen(str);
50 for(int i = 0; i < len; ++i) a[i] = str[i];
51 a[len] = 1;
52 da(a, sa, len+1, 555);
53 calheight(a, sa, len);
54 int sum = 0;
55 for(int i = 1; i <= len; ++i){
56 sum = sum + len-sa[i]-height[i];
57 }
58 printf("%d\n", sum);
59 }
60 return 0;
61 }
--------------------------------------------------------
到今天才是1/8的男人
http://poj.org/problem?id=1743（楼教主的男人八题之一）
求不重叠的最长公共串
1 #include <iostream>
2 #include <string.h>
3 #include <stdio.h>
4 using namespace std;
5
6 const int maxn = 20009;
7 int a[maxn], sa[maxn], len;
8 char str[maxn];
9
10 int wa[maxn], wb[maxn], wv[maxn], wws[maxn];
11 int cmp(int *r, int a, int b, int l){
12 return r[a] == r[b] && r[a+l] == r[b+l];
13 }
14 //sa from 0 -> n-1
15 void da(int *r, int *sa, int n, int m){
16 int i, j, p, *x = wa, *y = wb, *t;
17 for(i = 0; i < m; i++) wws[i] = 0;
18 for(i = 0; i < n; i++) wws[x[i] = r[i]]++;
19 for(i = 1; i < m; i++) wws[i] += wws[i-1];
20 for(i = n-1; i >= 0; i--) sa[--wws[x[i]]] = i;
21 for(j = 1,p = 1; p < n; j *= 2, m = p){
22 for(p = 0,i = n-j; i<n;i++) y[p++] = i;
23 for(i = 0; i < n; i++) if(sa[i] >= j) y[p++] = sa[i]-j;
24 for(i = 0; i < n; i++) wv[i] = x[y[i]];
25 for(i = 0; i < m; i++) wws[i] = 0;
26 for(i = 0; i < n; i++) wws[wv[i]] ++;
27 for(i = 1; i < m; i++) wws[i] += wws[i-1];
28 for(i = n-1; i >= 0; i--) sa[--wws[wv[i]]] = y[i];
29 for(t = x, x = y, y = t, p = 1, x[sa[0]] = 0, i = 1; i < n; i++)
30 x[sa[i]] = cmp(y,sa[i-1], sa[i],j) ? p-1 : p++;
31 }
32 return ;
33 }
34
35 int Rank[maxn], height[maxn];
36 void calheight(int *r, int *sa, int n){
37 int i, j, k = 0;
38 for(i = 1; i <= n; i++) Rank[sa[i]] = i;
39 for(i = 0; i < n; height[Rank[i++]] = k)
40 for(k ? k-- : 0, j = sa[Rank[i]-1]; r[i+k] == r[j+k]; k++);
41 return;
42 }
43
44 bool ok(int x, int len){
45 int mi = sa[1], ma = sa[1];
46 for(int i = 2; i <= len; ++i){
47 if(height[i] < x) mi = ma = sa[i];
48 else{
49 if(sa[i] < mi) mi = sa[i];
50 if(sa[i] > ma) ma = sa[i];
51 //不能 ma-mi >= x，同样a映射的是差值若相等必重叠
52 if(ma - mi > x) return true;
53 }
54 }
55 return false;
56 }
57
58 int main(){
59 int n, x, y;
60 while(~scanf("%d", &n) && n){
61 //n个数，a[i] 存储a[i+1]-a[i]的值 a的范围是 0 -> n-1
62 scanf("%d", &x);
63 for(int i = 0; i < n-1; ++i) scanf("%d", &y), a[i] = y-x+99, x = y;
64 a[n-1] = 0, n--;
65 da(a, sa, n+1, 555);
66 calheight(a, sa, n);
67 int l = 0, r = n, t = 0;
68 while(l <= r){
69 int mid = (l+r)>>1;
70 if(ok(mid, n)) t = mid, l = mid+1;
71 else r = mid-1;
72 }
73 //因为a为实际数组的差值，所以用四来判断
74 printf("%d\n", t >= 4 ? t+1 : 0);
75 }
76 return 0;
77 }
--------------------------------------------------------
uva11107
输入n个DNA序列，求出长度最大的字符串，使得它在超过一半的DNA序列中连续出现，若有多解按照字典序从小到大输出
1 #include<cstdio>
2 #include<cstring>
3 #include<algorithm>
4 using namespace std;
5
6 const int maxn = 1001 * 100 + 10;
7
8 struct SuffixArray {
9 int s[maxn]; // 原始字符数组（最后一个字符应必须是0，而前面的字符必须非0）
10 int sa[maxn]; // 后缀数组
11 int rank[maxn]; // 名次数组. rank[0]一定是n-1，即最后一个字符
12 int height[maxn]; // height数组
13 int t[maxn], t2[maxn], c[maxn]; // 辅助数组
14 int n; // 字符个数
15
16 void clear() { n = 0; memset(sa, 0, sizeof(sa)); }
17
18 // m为最大字符值加1。调用之前需设置好s和n
19 void build_sa(int m) {
20 int i, *x = t, *y = t2;
21 for(i = 0; i < m; i++) c[i] = 0;
22 for(i = 0; i < n; i++) c[x[i] = s[i]]++;
23 for(i = 1; i < m; i++) c[i] += c[i-1];
24 for(i = n-1; i >= 0; i--) sa[--c[x[i]]] = i;
25 for(int k = 1; k <= n; k <<= 1) {
26 int p = 0;
27 for(i = n-k; i < n; i++) y[p++] = i;
28 for(i = 0; i < n; i++) if(sa[i] >= k) y[p++] = sa[i]-k;
29 for(i = 0; i < m; i++) c[i] = 0;
30 for(i = 0; i < n; i++) c[x[y[i]]]++;
31 for(i = 0; i < m; i++) c[i] += c[i-1];
32 for(i = n-1; i >= 0; i--) sa[--c[x[y[i]]]] = y[i];
33 swap(x, y);
34 p = 1; x[sa[0]] = 0;
35 for(i = 1; i < n; i++)
36 x[sa[i]] = y[sa[i-1]]==y[sa[i]] && y[sa[i-1]+k]==y[sa[i]+k] ? p-1 : p++;
37 if(p >= n) break;
38 m = p;
39 }
40 }
41
42 void build_height() {
43 int i, j, k = 0;
44 for(i = 0; i < n; i++) rank[sa[i]] = i;
45 for(i = 0; i < n; i++) {
46 if(k) k--;
47 int j = sa[rank[i]-1];
48 while(s[i+k] == s[j+k]) k++;
49 height[rank[i]] = k;
50 }
51 }
52 };
53
54 const int maxc = 100 + 10; // 串的个数
55 const int maxl = 1000 + 10; // 每个串的长度
56
57 SuffixArray sa;
58 int n;
59 char word[maxl];
60 int idx[maxn];
61 int flag[maxc];
62
63 // 子串[L,R) 是否符合要求
64 bool good(int L, int R) {
65 memset(flag, 0, sizeof(flag));
66 if(R - L <= n/2) return false;
67 int cnt = 0;
68 for(int i = L; i < R; i++) {
69 int x = idx[sa.sa[i]];
70 if(x != n && !flag[x]) { flag[x] = 1; cnt++; }
71 }
72 return cnt > n/2;
73 }
74
75 void print_sub(int L, int R) {
76 for(int i = L; i < R; i++)
77 printf("%c", sa.s[i] - 1 + ‘a‘);
78 printf("\n");
79 }
80
81 bool print_solutions(int len, bool print) {
82 int L = 0;
83 for(int R = 1; R <= sa.n; R++) {
84 if(R == sa.n || sa.height[R] < len) { // 新开一段
85 if(good(L, R)) {
86 if(print) print_sub(sa.sa[L], sa.sa[L] + len); else return true;
87 }
88 L = R;
89 }
90 }
91 return false;
92 }
93
94 void solve(int maxlen) {
95 if(!print_solutions(1, false))
96 printf("?\n");
97 else {
98 int L = 1, R = maxlen, M;
99 while(L < R) {
100 M = L + (R-L+1)/2;
101 if(print_solutions(M, false)) L = M;
102 else R = M-1;
103 }
104 print_solutions(L, true);
105 }
106 }
107
108 // 给字符串加上一个字符，属于字符串i
109 void add(int ch, int i) {
110 idx[sa.n] = i;
111 sa.s[sa.n++] = ch;
112 }
113
114 int main() {
115 int kase = 0;
116 while(scanf("%d", &n) == 1 && n) {
117 if(kase++ > 0) printf("\n");
118 int maxlen = 0;
119 sa.clear();
120 for(int i = 0; i < n; i++) {
121 scanf("%s", word);
122 int sz = strlen(word);
123 maxlen = max(maxlen, sz);
124 for(int j = 0; j < sz; j++)
125 add(word[j] - ‘a‘ + 1, i);
126 add(100 + i, n); // 结束字符
127 }
128 add(0, n);
129
130 if(n == 1) printf("%s\n", word);
131 else {
132 sa.build_sa(100 + n);
133 sa.build_height();
134 solve(maxlen);
135 }
136 }
137 return 0;
138 }
----------------------------------------------------------------
http://acm.timus.ru/problem.aspx?space=1&num=1297
求最长回文串(当然可以用o（n）的算法实现)
1 #include <stdio.h>
2 #include <string.h>
3 #define maxn 2002
4
5 int wa[maxn],wb[maxn],wv[maxn],ws[maxn];
6
7 int cmp(int *r,int a,int b,int l){
8 return r[a]==r[b]&&r[a+l]==r[b+l];
9 }
10
11 void da(int *r,int *sa,int n,int m){
12 int i,j,p,*x=wa,*y=wb,*t;
13 for(i=0;i<m;i++) ws[i]=0;
14 for(i=0;i<n;i++) ws[x[i]=r[i]]++;
15 for(i=1;i<m;i++) ws[i]+=ws[i-1];
16 for(i=n-1;i>=0;i--) sa[--ws[x[i]]]=i;
17 for(j=1,p=1;p<n;j*=2,m=p){
18 for(p=0,i=n-j;i<n;i++) y[p++]=i;
19 for(i=0;i<n;i++) if(sa[i]>=j) y[p++]=sa[i]-j;
20 for(i=0;i<n;i++) wv[i]=x[y[i]];
21 for(i=0;i<m;i++) ws[i]=0;
22 for(i=0;i<n;i++) ws[wv[i]]++;
23 for(i=1;i<m;i++) ws[i]+=ws[i-1];
24 for(i=n-1;i>=0;i--) sa[--ws[wv[i]]]=y[i];
25 for(t=x,x=y,y=t,p=1,x[sa[0]]=0,i=1;i<n;i++)
26 x[sa[i]]=cmp(y,sa[i-1],sa[i],j)?p-1:p++;
27 }
28 return;
29 }
30
31 int rank[maxn],height[maxn];
32 void calheight(int *r,int *sa,int n){
33 int i,j,k=0;
34 for(i=1;i<=n;i++) rank[sa[i]]=i;
35 for(i=0;i<n;height[rank[i++]]=k)
36 for(k?k--:0,j=sa[rank[i]-1];r[i+k]==r[j+k];k++);
37 return;
38 }
39
40 int RMQ[maxn];
41 int mm[maxn];
42 int best[20][maxn];
43 void initRMQ(int n){
44 int i,j,a,b;
45 for(mm[0]=-1,i=1;i<=n;i++)
46 mm[i]=((i&(i-1))==0)?mm[i-1]+1:mm[i-1];
47 for(i=1;i<=n;i++) best[0][i]=i;
48 for(i=1;i<=mm[n];i++)
49 for(j=1;j<=n+1-(1<<i);j++){
50 a=best[i-1][j];
51 b=best[i-1][j+(1<<(i-1))];
52 if(RMQ[a]<RMQ[b]) best[i][j]=a;
53 else best[i][j]=b;
54 }
55 return;
56 }
57
58 int askRMQ(int a,int b){
59 int t;
60 t=mm[b-a+1];b-=(1<<t)-1;
61 a=best[t][a];b=best[t][b];
62 return RMQ[a]<RMQ[b]?a:b;
63 }
64
65 int lcp(int a,int b){
66 int t;
67 a=rank[a];b=rank[b];
68 if(a>b) {t=a;a=b;b=t;}
69 return(height[askRMQ(a+1,b)]);
70 }
71
72 char st[maxn];
73 int r[maxn],sa[maxn];
74
75 int main(){
76 int i,n,len,k,ans=0,w;
77 scanf("%s",st);
78 len=strlen(st);
79 for(i=0;i<len;i++) r[i]=st[i];
80 r[len]=1;
81 for(i=0;i<len;i++) r[i+len+1]=st[len-1-i];
82 n=len+len+1;
83 r[n]=0;
84 da(r,sa,n+1,128);
85 calheight(r,sa,n);
86 for(i=1;i<=n;i++) RMQ[i]=height[i];
87 initRMQ(n);
88 for(i=0;i<len;i++){
89 k=lcp(i,n-i);
90 if(k*2>ans) ans=k*2,w=i-k;
91 k=lcp(i,n-i-1);
92 if(k*2-1>ans) ans=k*2-1,w=i-k+1;
93 }
94 st[w+ans]=0;
95 printf("%s\n",st+w);
96 return 0;
97 }
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http://poj.org/problem?id=2406
求字符串的最长循环节（KMP裸题），这里看后缀数组解法
1 #include <iostream>
2 #include <stdio.h>
3 #include <string.h>
4 using namespace std;
5
6 #define maxn 1000010
7 #define F(x) ((x)/3+((x)%3 == 1?0:tb))
8 #define G(x) ((x)<tb ? (x)*3+1 : ((x)-tb)*3+2)
9
10 int wa[maxn], wb[maxn], wv[maxn], wss[maxn];
11 int c0(int *r, int a, int b) {
12 return r[a] == r[b] && r[a+1] == r[b+1] && r[a+2] == r[b+2];
13 }
14 int c12(int k, int *r, int a, int b) {
15 if(k == 2) return r[a] < r[b] || r[a] == r[b] && c12(1, r, a+1, b+1);
16 else return r[a] < r[b] || r[a] == r[b] && wv[a+1] < wv[b+1];
17 }
18
19 void sort(int *r, int *a, int *b, int n, int m) {
20 int i;
21 for(i = 0; i < n; i++) wv[i] = r[a[i]];
22 for(i = 0; i < m; i++) wss[i] = 0;
23 for(i = 0; i < n; i++) wss[wv[i]] ++;
24 for(i = 1; i < m; i++) wss[i] += wss[i-1];
25 for(i = n-1; i >= 0; i--) b[--wss[wv[i]]] = a[i];
26 return;
27 }
28
29 void dc3(int *r, int *sa, int n, int m) {
30 int i, j, *rn = r+n, *san = sa+n, ta = 0, tb = (n+1)/3, tbc=0, p;
31 r[n] = r[n+1] = 0;
32 for(i = 0; i < n; i++) if(i%3 != 0) wa[tbc++] = i;
33 sort(r+2, wa, wb, tbc, m);
34 sort(r+1, wb, wa, tbc, m);
35 sort(r, wa, wb, tbc, m);
36 for(p = 1, rn[F(wb[0])] = 0, i = 1; i < tbc; i++)
37 rn[F(wb[i])] = c0(r, wb[i-1], wb[i]) ? p-1 : p++;
38 if(p < tbc) dc3(rn, san, tbc, p);
39 else for(i = 0; i < tbc; i++) san[rn[i]] = i;
40 for(i = 0; i < tbc; i++) if(san[i] < tb) wb[ta++] = san[i]*3;
41 if(n%3 == 1) wb[ta++] = n-1;
42 sort(r, wb, wa, ta, m);
43 for(i = 0; i < tbc; i++) wv[wb[i]=G(san[i])] = i;
44 for(i = 0, j = 0, p = 0; i < ta && j < tbc; p++)
45 sa[p] = c12(wb[j]%3, r, wa[i], wb[j]) ? wa[i++] : wb[j++];
46 for(; i < ta; p++) sa[p] = wa[i++];
47 for(; j < tbc; p++) sa[p] = wb[j++];
48 return;
49 }
50
51 int rk[3*maxn], height[3*maxn];
52 void calheight(int *r, int *sa, int n) {
53 int i, j, k = 0;
54 for(i = 1; i <= n; i++) rk[sa[i]] = i;
55 for(i = 0; i < n; height[rk[i++]] = k)
56 for(k ? k-- : 0, j = sa[rk[i]-1]; r[i+k] == r[j+k]; k++);
57 return;
58 }
59
60 char str[maxn];
61 int r[maxn], num[maxn], sa[3*maxn];
62
63 int main(){
64 while(scanf("%s", str) && str[0] != ‘.‘){
65 int len = strlen(str);
66 for(int i = 0; i < len; ++i){
67 r[i] = str[i]-‘a‘+2;
68 }
69 r[len] = 1;
70 dc3(r, sa, len+1, 30);
71 calheight(r, sa, len);
72 int x = rk[0], ma = maxn;
73 num[x] = len;
74 for(int i = x-1; i >= 0; --i){
75 num[i] = min(ma, height[i+1]);
76 //cout<<"^^^"<<endl;
77 }
78 ma = maxn;
79 for(int i = x+1; i <= len; ++i){
80 num[i] = min(ma, height[i]);
81 }
82 int k = 1;
83 for(int i = 1; i <= len; ++i){
84 //cout << i << "&&" << num[i] << endl;
85 if(len%i == 0 && num[rk[i]] == len-i){
86 printf("%d\n", len/i);
87 break;
88 }
89 }
90 }
91 }
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只有不断学习才能进步！
原文地址：https://www.cnblogs.com/wenbao/p/7401180.html
时间： 2024-10-09 00:44:17