标签：splay stream -- 算法实现 controls ret kmp裸题 using i++
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];
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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 }
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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 }
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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 }
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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！
标签：splay stream -- 算法实现 controls ret kmp裸题 using i++
原文地址：https://www.cnblogs.com/wenbao/p/7401180.html
