POJ 1743
Meaning of the questions:
There are N (1 <= N <= 20000) of the sequence of notes of a piece of music is represented, each note is an integer in the range of 1 ~ 88, it is to find a duplicate topics. "Subject" is a substring of the entire sequence of notes, it needs to meet the following conditions:
1. The length of at least five notes.
2. repeated in the song. (Possibly transpose, "transpose" means relating to each note sequence is the same plus or minus an integer value)
3. recurring theme in the same original sequence not overlap.
question type:
May not overlap the longest repeated substring
analysis:
Transpose problem because, the difference can be an array of adjacent notes do not overlap the longest repeated substring
Then converted to solving common suffix array type
(Excerpt from the National Team thesis Lo Sui Qian):
First binary answer, the subject into Entscheidungsproblem: determining whether there are two sub-strings of length k is the same,
And do not overlap. The key to solving this problem is the use of an array of height. Suffix sorted into several groups,
Wherein the value of the height between each suffix is not less than k. For example, the string "aabaaaab"
When k = 2, suffix divided into four groups, as shown in FIG.
Readily seen, it promises to be the longest common prefix is not less than two suffixes k must be in the same group.
Then, for each suffix, only determines the difference between the maximum and minimum values for each suffix sa is not less than k.
If you have a group meet, then there is, or do not exist. Time complexity of the whole approach is O (nlogn).
1 #include <cstdio>
2 #include <cstring>
3 #include <queue>
4 #include <cmath>
5 #include <algorithm>
6 #include <set>
7 #include <iostream>
8 #include <map>
9 #include <stack>
10 #include <string>
11 #include <time.h>
12 #include <vector>
13 #define pi acos(-1.0)
14 #define eps 1e-9
15 #define fi first
16 #define se second
17 #define rtl rt<<1
18 #define rtr rt<<1|1
19 #define bug printf("******\n")
20 #define mem(a,b) memset(a,b,sizeof(a))
21 #define name2str(x) #x
22 #define fuck(x) cout<<#x" = "<<x<<endl
23 #define f(a) a*a
24 #define sf(n) scanf("%d", &n)
25 #define sff(a,b) scanf("%d %d", &a, &b)
26 #define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
27 #define sffff(a,b,c,d) scanf("%d %d %d %d", &a, &b, &c, &d)
28 #define pf printf
29 #define FRE(i,a,b) for(i = a; i <= b; i++)
30 #define FREE(i,a,b) for(i = a; i >= b; i--)
31 #define FRL(i,a,b) for(i = a; i < b; i++)+
32 #define FRLL(i,a,b) for(i = a; i > b; i--)
33 #define FIN freopen("data.txt","r",stdin)
34 #define gcd(a,b) __gcd(a,b)
35 #define lowbit(x) x&-x
36 #define rep(i,a,b) for(int i=a;i<b;++i)
37 #define per(i,a,b) for(int i=a-1;i>=b;--i)
38
39 using namespace std;
40 typedef long long LL;
41 typedef unsigned long long ULL;
42 const int maxn = 1e5 + 7;
43 const int maxm = 8e6 + 10;
44 const int INF = 0x3f3f3f3f;
45 const int mod = 10007;
46
47 //rnk从0开始
48 //sa从1开始,因为最后一个字符(最小的)排在第0位
49 //height从1开始,因为表示的是sa[i - 1]和sa[i]
50 //倍增算法 O(nlogn)
51 int wa[maxn], wb[maxn], wv[maxn], ws_[maxn];
52 int Rank[maxn], height[maxn], sa[maxn], s[maxn];
53 int n;
54 //Suffix函数的参数m代表字符串中字符的取值范围,是基数排序的一个参数,如果原序列都是字母可以直接取128,如果原序列本身都是整数的话,则m可以取比最大的整数大1的值
55 //待排序的字符串放在r数组中,从r[0]到r[n-1]，长度为n
56 //为了方便比较大小,可以在字符串后面添加一个字符,这个字符没有在前面的字符中出现过,而且比前面的字符都要小
57 //同上,为了函数操作的方便,约定除r[n-1]外所有的r[i]都大于0,r[n-1]=0
58 //函数结束后,结果放在sa数组中,从sa[0]到sa[n-1]
59 void Suffix ( int *r, int *sa, int n, int m ) {
60 int i, j, k, *x = wa, *y = wb, *t;
61 //对长度为1的字符串排序
62 //一般来说,在字符串的题目中,r的最大值不会很大,所以这里使用了基数排序
63 //如果r的最大值很大,那么把这段代码改成快速排序
64 for ( i = 0; i < m; ++i ) ws_[i] = 0;
65 for ( i = 0; i < n; ++i ) ws_[x[i] = r[i]]++; //统计字符的个数
66 for ( i = 1; i < m; ++i ) ws_[i] += ws_[i - 1]; //统计不大于字符i的字符个数
67 for ( i = n - 1; i >= 0; --i ) sa[--ws_[x[i]]] = i; //计算字符排名
68 //基数排序
69 //x数组保存的值相当于是rank值
70 for ( j = 1, k = 1; k < n; j *= 2, m = k ) {
71 //j是当前字符串的长度,数组y保存的是对第二关键字排序的结果
72 //第二关键字排序
73 for ( k = 0, i = n - j; i < n; ++i ) y[k++] = i; //第二关键字为0的排在前面
74 for ( i = 0; i < n; ++i ) if ( sa[i] >= j ) y[k++] = sa[i] - j; //长度为j的子串sa[i]应该是长度为2 * j的子串sa[i] - j的后缀（第二关键字）,对所有的长度为2 * j的子串根据第二关键字来排序
75 for ( i = 0; i < n; ++i ) wv[i] = x[y[i]]; //提取第一关键字
76 //按第一关键字排序 (原理同对长度为1的字符串排序)
77 for ( i = 0; i < m; ++i ) ws_[i] = 0;
78 for ( i = 0; i < n; ++i ) ws_[wv[i]]++;
79 for ( i = 1; i < m; ++i ) ws_[i] += ws_[i - 1];
80 for ( i = n - 1; i >= 0; --i ) sa[--ws_[wv[i]]] = y[i]; //按第一关键字,计算出了长度为2 * j的子串排名情况
81 //此时数组x是长度为j的子串的排名情况,数组y仍是根据第二关键字排序后的结果
82 //计算长度为2 * j的子串的排名情况,保存到数组x
83 t = x;
84 x = y;
85 y = t;
86 for ( x[sa[0]] = 0, i = k = 1; i < n; ++i )
87 x[sa[i]] = ( y[sa[i - 1]] == y[sa[i]] && y[sa[i - 1] + j] == y[sa[i] + j] ) ? k - 1 : k++;
88 //若长度为2 * j的子串sa[i]与sa[i - 1]完全相同,则他们有相同的排名
89 }
90 }
91 void calheight ( int *r, int *sa, int n ) {
92 int i, j, k = 0;
93 for ( i = 1; i <= n; i++ ) Rank[sa[i]] = i;
94 for ( i = 0; i < n; height[Rank[i++]] = k )
95 for ( k ? k-- : 0, j = sa[Rank[i] - 1]; r[i + k] == r[j + k]; k++ );
96 }
97 bool judge ( int c ) {
98 int Max = sa[0], Min = sa[0];
99 for ( int i = 1; i <= n; i++ ) {
100 if ( height[i] >= c )
101 Max = max ( Max, sa[i] ), Min = min ( Min, sa[i] );
102 else
103 Max = sa[i], Min = sa[i];
104 if ( Max - Min >= c + 1 )
105 return true;
106 }
107 return false;
108 }
109 int main() {
110 while ( sf ( n ) && n ) {
111 int maxx = 0;
112 for ( int i = 0; i < n ; i++ ) {
113 sf ( s[i] );
114 if ( i ) s[i - 1] = s[i] - s[i - 1] + 88, maxx = max ( maxx, s[i - 1] );
115 }
116 s[n-1] = 0;
117 n--;
118 Suffix ( s, sa, n + 1, maxx + 1 );
119 calheight ( s, sa, n );
120 int low = 0, high = n, ans = 0;
121 while ( low <= high ) {
122 int mid = ( low + high ) / 2;
123 if ( judge ( mid ) ) {
124 low = mid + 1;
125 ans = mid;
126 } else high = mid - 1;
127 }
128 if ( ans < 4 ) printf ( "0\n" );
129 else printf ( "%d\n", ans + 1 );
130 }
131 return 0;
132 }
POJ 3261 Milk Patterns
题意：
给出一个字符串，求至少出现k次的可重叠的最长子串的长度
(摘自罗穗骞的国家集训队论文):
算法分析： 这题的做法和上一题差不多，也是先二分答案，然后将后缀分成若干组。
不同的是，这里要判断的是有没有一个组的后缀个数不小于 k。
如果有，那么存在 k 个相同的子串满足条件，否则不存在。这个做法的时间复杂度为 O(nlogn)。
1 #include <cstdio>
2 #include <cstring>
3 #include <queue>
4 #include <cmath>
5 #include <algorithm>
6 #include <set>
7 #include <iostream>
8 #include <map>
9 #include <stack>
10 #include <string>
11 #include <time.h>
12 #include <vector>
13 #define pi acos(-1.0)
14 #define eps 1e-9
15 #define fi first
16 #define se second
17 #define rtl rt<<1
18 #define rtr rt<<1|1
19 #define bug printf("******\n")
20 #define mem(a,b) memset(a,b,sizeof(a))
21 #define name2str(x) #x
22 #define fuck(x) cout<<#x" = "<<x<<endl
23 #define f(a) a*a
24 #define sf(n) scanf("%d", &n)
25 #define sff(a,b) scanf("%d %d", &a, &b)
26 #define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
27 #define sffff(a,b,c,d) scanf("%d %d %d %d", &a, &b, &c, &d)
28 #define pf printf
29 #define FRE(i,a,b) for(i = a; i <= b; i++)
30 #define FREE(i,a,b) for(i = a; i >= b; i--)
31 #define FRL(i,a,b) for(i = a; i < b; i++)+
32 #define FRLL(i,a,b) for(i = a; i > b; i--)
33 #define FIN freopen("data.txt","r",stdin)
34 #define gcd(a,b) __gcd(a,b)
35 #define lowbit(x) x&-x
36 #define rep(i,a,b) for(int i=a;i<b;++i)
37 #define per(i,a,b) for(int i=a-1;i>=b;--i)
38
39 using namespace std;
40 typedef long long LL;
41 typedef unsigned long long ULL;
42 const int maxn = 1e5 + 7;
43 const int maxm = 8e6 + 10;
44 const int INF = 0x3f3f3f3f;
45 const int mod = 10007;
46
47 //rnk从0开始
48 //sa从1开始,因为最后一个字符(最小的)排在第0位
49 //height从1开始,因为表示的是sa[i - 1]和sa[i]
50 //倍增算法 O(nlogn)
51 int wa[maxn], wb[maxn], wv[maxn], ws_[maxn];
52 int Rank[maxn], height[maxn], sa[maxn], s[maxn];
53 int n, K;
54 //Suffix函数的参数m代表字符串中字符的取值范围,是基数排序的一个参数,如果原序列都是字母可以直接取128,如果原序列本身都是整数的话,则m可以取比最大的整数大1的值
55 //待排序的字符串放在r数组中,从r[0]到r[n-1]，长度为n
56 //为了方便比较大小,可以在字符串后面添加一个字符,这个字符没有在前面的字符中出现过,而且比前面的字符都要小
57 //同上,为了函数操作的方便,约定除r[n-1]外所有的r[i]都大于0,r[n-1]=0
58 //函数结束后,结果放在sa数组中,从sa[0]到sa[n-1]
59 void Suffix ( int *r, int *sa, int n, int m ) {
60 int i, j, k, *x = wa, *y = wb, *t;
61 //对长度为1的字符串排序
62 //一般来说,在字符串的题目中,r的最大值不会很大,所以这里使用了基数排序
63 //如果r的最大值很大,那么把这段代码改成快速排序
64 for ( i = 0; i < m; ++i ) ws_[i] = 0;
65 for ( i = 0; i < n; ++i ) ws_[x[i] = r[i]]++; //统计字符的个数
66 for ( i = 1; i < m; ++i ) ws_[i] += ws_[i - 1]; //统计不大于字符i的字符个数
67 for ( i = n - 1; i >= 0; --i ) sa[--ws_[x[i]]] = i; //计算字符排名
68 //基数排序
69 //x数组保存的值相当于是rank值
70 for ( j = 1, k = 1; k < n; j *= 2, m = k ) {
71 //j是当前字符串的长度,数组y保存的是对第二关键字排序的结果
72 //第二关键字排序
73 for ( k = 0, i = n - j; i < n; ++i ) y[k++] = i; //第二关键字为0的排在前面
74 for ( i = 0; i < n; ++i ) if ( sa[i] >= j ) y[k++] = sa[i] - j; //长度为j的子串sa[i]应该是长度为2 * j的子串sa[i] - j的后缀（第二关键字）,对所有的长度为2 * j的子串根据第二关键字来排序
75 for ( i = 0; i < n; ++i ) wv[i] = x[y[i]]; //提取第一关键字
76 //按第一关键字排序 (原理同对长度为1的字符串排序)
77 for ( i = 0; i < m; ++i ) ws_[i] = 0;
78 for ( i = 0; i < n; ++i ) ws_[wv[i]]++;
79 for ( i = 1; i < m; ++i ) ws_[i] += ws_[i - 1];
80 for ( i = n - 1; i >= 0; --i ) sa[--ws_[wv[i]]] = y[i]; //按第一关键字,计算出了长度为2 * j的子串排名情况
81 //此时数组x是长度为j的子串的排名情况,数组y仍是根据第二关键字排序后的结果
82 //计算长度为2 * j的子串的排名情况,保存到数组x
83 t = x;
84 x = y;
85 y = t;
86 for ( x[sa[0]] = 0, i = k = 1; i < n; ++i )
87 x[sa[i]] = ( y[sa[i - 1]] == y[sa[i]] && y[sa[i - 1] + j] == y[sa[i] + j] ) ? k - 1 : k++;
88 //若长度为2 * j的子串sa[i]与sa[i - 1]完全相同,则他们有相同的排名
89 }
90 }
91 void calheight ( int *r, int *sa, int n ) {
92 int i, j, k = 0;
93 for ( i = 1; i <= n; i++ ) Rank[sa[i]] = i;
94 for ( i = 0; i < n; height[Rank[i++]] = k )
95 for ( k ? k-- : 0, j = sa[Rank[i] - 1]; r[i + k] == r[j + k]; k++ );
96 }
97 bool judge ( int mid ) {
98 int cnt = 0;
99 for ( int i = 1; i <= n; i++ ) {
100 if ( height[i] >= mid ) cnt++;
101 else cnt = 0;
102 if ( cnt+1 >= K ) return true;
103 }
104 return false;
105 }
106 int main() {
107 while ( ~sff ( n, K ) ) {
108 int maxx = 0;
109 for ( int i = 0; i < n ; i++ ) sf ( s[i] ), maxx = max ( maxx, s[i] );
110 s[n] = 0;
111 Suffix ( s, sa, n + 1, maxx + 1 );
112 calheight ( s, sa, n );
113 int low = 0, high = n, ans = 0;
114 while ( low <= high ) {
115 int mid = ( low + high ) / 2;
116 if ( judge ( mid ) ) {
117 low = mid + 1;
118 ans = mid;
119 } else high = mid - 1;
120 }
121 printf ( "%d\n", ans );
122 }
123 return 0;
124 }