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.

#include<cstdlib>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cassert>
using namespace std;
const int maxn = (int)1.01e6,sigma = 26;
char str[maxn];
int Min[maxn * 2],Max[maxn * 2],id[maxn * 2];
int cmq(int,int);
struct Sam{
    int ch[maxn * 2][sigma],par[maxn * 2],stp[maxn * 2];
    int sz,last;
    void init(){
        sz = last = 1;
    }
    void ext(int c){
        stp[++sz] = stp[last] + 1;
        int p = last, np = sz;
        for(; !ch[p][c]; p = par[p]) ch[p][c] = np;
        if(p == 0) par[np] = 1;
        else{
            int q = ch[p][c];
            if(stp[q] != stp[p] + 1){
                stp[++sz] = stp[p] + 1;
                int nq = sz;
                memcpy(ch[nq],ch[q],sizeof(ch[q]));
                par[nq] = par[q];
                par[q] = par[np] = nq;
                for(; ch[p][c] == q; p = par[p]) ch[p][c] = nq;
            }
            else par[np] = q;
        }
        last = np;
    }
    void ins(char *pt){
        int i;
        init();
        for(i = 0; pt[i]; ++i) ext(pt[i] - 'a');
    }
    void prep(){
        int i;
        for(i = 1; i <= sz; ++i) Min[i] = stp[i];
        for(i = 1; i <= sz; ++i) id[i] = i;
        sort(id + 1, id + sz + 1,cmq); 
    }
    void cmp(char *pt){
        int i,x = 1,cnt = 0;
        fill(Max, Max + sz + 2, 0);
        for(i = 0; pt[i]; ++i){
            if(ch[x][pt[i] - 'a']){
                x = ch[x][pt[i] - 'a'];
                Max[x] = max(Max[x],++cnt);
            }
            else{
                while(x != 1 && !ch[x][pt[i] - 'a']) x = par[x], cnt = stp[x];
                if(ch[x][pt[i] - 'a']) x = ch[x][pt[i] - 'a'],Max[x] = max(Max[x],++cnt);
            }
        }        
        for(i = 1; i <= sz; ++i){
            if(id[i] != 1 && Min[par[id[i]]] && Max[id[i]])
                assert(Min[par[id[i]]] + 1 <= Max[id[i]]);
            Min[id[i]] = min(Min[id[i]], Max[id[i]]);            
            Max[par[id[i]]] = max(Max[par[id[i]]], Max[id[i]]);
        }
    }
}sam;
int cmq(int x,int y){
    return sam.stp[x] > sam.stp[y];
}
int main()
{
    freopen("substr.in","r",stdin);
    freopen("substr.out","w",stdout);
    scanf("%s",str);
    sam.ins(str);
    sam.prep();
    while(scanf("%s",str) != EOF) 
        sam.cmp(str);
    printf("%d
",*max_element(Min + 1, Min + sam.sz + 1));
    return 0;
}