spoj 1812
1812. Longest Common Substring II
Problem code: LCS2
A string is finite sequence of characters over a non-empty finite set Σ.
In this problem, Σ is the set of lowercase letters.
Substring, also called factor, is a consecutive sequence of characters occurrences at least once in a string.
Now your task is a bit harder, for some given strings, find the length of the longest common substring of them.
Here common substring means a substring of two or more strings.
Input
The input contains at most 10 lines, each line consists of no more than 100000 lowercase letters, representing a string.
Output
The length of the longest common substring. If such string doesn't exist, print "0" instead.
Example
Input:
alsdfkjfjkdsal
fdjskalajfkdsla
aaaajfaaaa
Output:
2
大意:
求出多个串的最长公共字串,输出长度.
分析:
我们这么想:
我们先拿两个串A,B匹配,然后在两个串的匹配过程中计算出后缀自动机SAM_A中某个节点x能在B中匹配的最长的长度.
那么这个问题就解决了.(我可以继续匹配C,D)最后取个min.
裸地解决这个问题是O(L2)的.
所以我们应用上spoj1811中的性质: parent _x 和 x 节点的最长公共部分是 max_parent_x.
那么我们只要利用孩子的匹配长度就能够直接推出parent的匹配长度. min(Min[par[x]], Max[x])
(Min[x] 表示从x节点开始在当前已经考虑的串中,从x节点出发能够拓展的最长的长度. Max[x] 表示在当前串内, 从x节点出发, 能够匹配的最长的长度.)
值得注意的是,当 Min[par[x]] != 0 && Max[x] != 0 时, Min[par[x]] + 1 <= Max[x].
所以,我们现在的关键在于面对 Max[x] == 0 的情况.
这种情况下,表明从x这个点无法进行拓展,所以自然表示par[x] 也无法进行拓展,因为par是x的极大后缀串.
所以我们迭代更新Min的值,最后取Min就ok了.
代码中加入了一个特判来验证我的猜想:
偷懒没有用"鸡排"....见谅.
1 #include<cstdlib> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<cassert> 6 using namespace std; 7 const int maxn = (int)1.01e6,sigma = 26; 8 char str[maxn]; 9 int Min[maxn * 2],Max[maxn * 2],id[maxn * 2]; 10 int cmq(int,int); 11 struct Sam{ 12 int ch[maxn * 2][sigma],par[maxn * 2],stp[maxn * 2]; 13 int sz,last; 14 void init(){ 15 sz = last = 1; 16 } 17 void ext(int c){ 18 stp[++sz] = stp[last] + 1; 19 int p = last, np = sz; 20 for(; !ch[p][c]; p = par[p]) ch[p][c] = np; 21 if(p == 0) par[np] = 1; 22 else{ 23 int q = ch[p][c]; 24 if(stp[q] != stp[p] + 1){ 25 stp[++sz] = stp[p] + 1; 26 int nq = sz; 27 memcpy(ch[nq],ch[q],sizeof(ch[q])); 28 par[nq] = par[q]; 29 par[q] = par[np] = nq; 30 for(; ch[p][c] == q; p = par[p]) ch[p][c] = nq; 31 } 32 else par[np] = q; 33 } 34 last = np; 35 } 36 void ins(char *pt){ 37 int i; 38 init(); 39 for(i = 0; pt[i]; ++i) ext(pt[i] - 'a'); 40 } 41 void prep(){ 42 int i; 43 for(i = 1; i <= sz; ++i) Min[i] = stp[i]; 44 for(i = 1; i <= sz; ++i) id[i] = i; 45 sort(id + 1, id + sz + 1,cmq); 46 } 47 void cmp(char *pt){ 48 int i,x = 1,cnt = 0; 49 fill(Max, Max + sz + 2, 0); 50 for(i = 0; pt[i]; ++i){ 51 if(ch[x][pt[i] - 'a']){ 52 x = ch[x][pt[i] - 'a']; 53 Max[x] = max(Max[x],++cnt); 54 } 55 else{ 56 while(x != 1 && !ch[x][pt[i] - 'a']) x = par[x], cnt = stp[x]; 57 if(ch[x][pt[i] - 'a']) x = ch[x][pt[i] - 'a'],Max[x] = max(Max[x],++cnt); 58 } 59 } 60 for(i = 1; i <= sz; ++i){ 61 if(id[i] != 1 && Min[par[id[i]]] && Max[id[i]]) 62 assert(Min[par[id[i]]] + 1 <= Max[id[i]]); 63 Min[id[i]] = min(Min[id[i]], Max[id[i]]); 64 Max[par[id[i]]] = max(Max[par[id[i]]], Max[id[i]]); 65 } 66 } 67 }sam; 68 int cmq(int x,int y){ 69 return sam.stp[x] > sam.stp[y]; 70 } 71 int main() 72 { 73 freopen("substr.in","r",stdin); 74 freopen("substr.out","w",stdout); 75 scanf("%s",str); 76 sam.ins(str); 77 sam.prep(); 78 while(scanf("%s",str) != EOF) 79 sam.cmp(str); 80 printf("%d\n",*max_element(Min + 1, Min + sam.sz + 1)); 81 return 0; 82 }
