SP1812 LCS2 - Longest Common Substring II

最新推荐文章于 2021-05-17 10:01:19 发布

转载最新推荐文章于 2021-05-17 10:01:19 发布 · 73 阅读

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原文链接：http://www.cnblogs.com/luoyibujue/p/10606097.html

本文介绍了一种基于AC自动机的字符串匹配算法，并提供了详细的实现代码。该算法通过构建一棵节点树来高效地处理多个模式串的匹配问题，在每次匹配过程中记录最长匹配长度，更新匹配结果。

能匹配上子串的节点对它的所有parent都有贡献
在树上转移即可

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<queue>
#define ll long long 
#define M 200010
using namespace std;

int read()
{
    int nm = 0, f = 1;
    char c = getchar();
    for(; !isdigit(c); c = getchar()) if(c == '-') f = -1;
    for(; isdigit(c); c = getchar()) nm = nm * 10 + c - '0';
    return nm * f;
}

int ch[M][26], len[M], fa[M], tim[M], a[M], cnt = 1, lst = 1, dp[M], ans[M];
char s[M];

void insert(int c)
{
    int p = ++cnt, f = lst;
    lst = p;
    len[p] = len[f] + 1;
    while(f && !ch[f][c]) ch[f][c] = p, f = fa[f];
    if(!f) fa[p] = 1;
    else
    {
        int q = ch[f][c];
        if(len[q] == len[f] + 1) fa[p] = q;
        else
        {
            int nq = ++cnt;
            fa[nq] = fa[q];
            memcpy(ch[nq], ch[q], sizeof(ch[nq]));
            len[nq] = len[f] + 1;
            fa[q] = fa[p] = nq;
            while(f && ch[f][c] == q) ch[f][c] = nq, f = fa[f];
        }
    }
}

int main()
{
    scanf("%s", s + 1);
    int l = strlen(s + 1);
    for(int i = 1; i <= l; i++) insert(s[i] - 'a');
    for(int i = 1; i <= cnt; i++) tim[len[i]]++;
    for(int i = 1; i <= cnt; i++) tim[i] += tim[i - 1];
    for(int i = 1; i <= cnt; i++) a[tim[len[i]]--] = i;
    for(int i = 1; i <= cnt; i++) ans[i] = len[i];
    while(scanf("%s", s + 1) != EOF)
    {
        memset(dp, 0, sizeof(dp));
        l = strlen(s + 1);
        int now = 1, tt = 0;
        for(int i = 1; i <= l; i++)
        {
            int c = s[i] - 'a';
            if(ch[now][c])
            {
                tt++, now = ch[now][c];
            }
            else
            {
                while(now && !ch[now][c]) now = fa[now];
                if(!now) now = 1, tt = 0;
                else tt = len[now] + 1, now = ch[now][c];
            }
            dp[now] = max(dp[now], tt);
        }
        for(int i = cnt; i >= 0; i--) dp[fa[i]] = max(dp[fa[i]], dp[i]);
        for(int i = 1; i <= cnt; i++) ans[i] = min(ans[i], dp[i]);
    }
    for(int i = 1; i <= cnt; i++) ans[i] = max(ans[i], ans[i - 1]);
    cout << ans[cnt] << "\n"; 
    return 0;
}

转载于:https://www.cnblogs.com/luoyibujue/p/10606097.html