SPOJ - LCS2 Longest Common Substring II

最新推荐文章于 2020-02-01 01:24:51 发布

Insist111

最新推荐文章于 2020-02-01 01:24:51 发布

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分类专栏：后缀自动机

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本文介绍了一种解决SPOJ平台上的LCS2问题的方法，通过构建后缀自动机来找出多个字符串间的最长公共子串。文章详细展示了如何初始化后缀自动机并逐步扩展以匹配新的字符串。

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SPOJ - LCS2

求多个字符串的最长公共子串。先构造第一个串的后缀自动机，定义一个nl表示当前字符串在该处与后缀自动机匹配的最大长度，ml用于保存所有字符串在该处都能匹配的长度，不断更新。

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
#include <bitset>
#define INF 0x3f3f3f3f
#define eps 1e-6
#define PI 3.1415926
#define mod 1000000009
#define base 2333
using namespace std;
typedef long long LL;
const int maxn = 1e5 + 10;
const int maxx = 1e3 + 10;
inline void splay(int &v) {
    v=0;char c=0;int p=1;
    while(c<'0' || c >'9'){if(c=='-')p=-1;c=getchar();}
    while(c>='0' && c<='9'){v=(v<<3)+(v<<1)+c-'0';c=getchar();}
    v*=p;
}
struct Node {
    Node *pre, *nxt[26];
    int step, nl, ml;
    void Clear() {
        step = nl = 0;
        pre = NULL;
        memset(nxt, NULL, sizeof(nxt));
    }
} *root, *last;
Node st[maxn<<1], *top[maxn<<1], *cur;
int cnt[maxn];
char str[15][maxn];

void init() {
    cur = st;
    root = last = cur++;
    root->Clear();
}

void extend(int w) {
    Node *np = cur++, *p = last;
    np->Clear();
    np->step = np->ml = p->step+1;
    while(p && !p->nxt[w])
        p->nxt[w] = np, p = p->pre;
    if(p == NULL)
        np->pre = root;
    else {
        Node *q = p->nxt[w];
        if(q->step == p->step+1)
            np->pre = q;
        else {
            Node *nq = cur++;
            nq->Clear();
            memcpy(nq->nxt, q->nxt, sizeof(q->nxt));
            nq->step = nq->ml = p->step+1;
            nq->pre = q->pre;
            np->pre = q->pre = nq;
            while(p && p->nxt[w] == q)
                p->nxt[w] = nq, p = p->pre;
        }
    }
    last = np;
}

void solve() {
    int k = 0;
    while(scanf("%s", str[k++]) != EOF) {}
    init();
    memset(cnt, 0, sizeof(cnt));
    int len = strlen(str[0]);
    for(int i = 0; i < len; i++)
        extend(str[0][i]-'a');
    for(Node *p = st; p != cur; p++)
        cnt[p->step]++;
    for(int i = 1; i <= len; i++)
        cnt[i] += cnt[i-1];
    for(Node *p = st; p != cur; p++)
        top[--cnt[p->step]] = p;
    int num = cur-st;
    for(int i = 1; i < k-1; i++) {
        int len1 = strlen(str[i]);
        Node *p = root;
        for(int j = 0, l = 0; j < len1; j++) {
            int x = str[i][j]-'a';
            if(p->nxt[x])
                l++, p = p->nxt[x];
            else {
                while(p && !p->nxt[x])
                    p = p->pre;
                if(p == NULL)
                    l = 0, p = root;
                else
                    l = p->step+1, p = p->nxt[x];
            }
            if(l > p->nl) p->nl = l;
        }
        for(int j = num-1; j >= 0; j--) {
            p = top[j];
            if(p->nl < p->ml)
                p->ml = p->nl;
            if(p->pre && p->pre->nl < p->nl)
                p->pre->nl = p->nl;
            p->nl = 0;

        }
    }
    int ans = 0;
    for(int i = 0; i < num; i++)
        ans = max(ans, top[i]->ml);
    printf("%d\n", ans);
}

int main() {
    //srand(time(NULL));
    //freopen("kingdom.in","r",stdin);
    //freopen("kingdom.out","w",stdout);
    solve();
}