SPOJ: LCS - Longest Common Substring-优快云博客

本文介绍使用后缀自动机(SAM)解决两个字符串的最长公共子串问题的方法。通过构建SAM并进行匹配来高效地找出最长公共子串。代码实现展示了如何构建SAM及匹配过程。

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题面

输入\(2\)个长度不大于\(250000\)的字符串，输出这\(2\)个字符串的最长公共子串。如果没有公共子串则输出\(0\)

Sol

一个串建立\(sam\)
另一个串在上面匹配

# include <bits/stdc++.h>
# define IL inline
# define RG register
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;

template <class Int>
IL void Input(RG Int &x){
    RG int z = 1; RG char c = getchar(); x = 0;
    for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
    for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
    x *= z;
}

const int maxn(5e5 + 5);

int n, trans[26][maxn], fa[maxn], len[maxn], tot = 1, last = 1, ans;
char s[maxn];

IL void Extend(RG int c){
    RG int np = ++tot, p = last; last = np;
    len[np] = len[p] + 1;
    while(p && !trans[c][p]) trans[c][p] = np, p = fa[p];
    if(!p) fa[np] = 1;
    else{
        RG int q = trans[c][p];
        if(len[q] == len[p] + 1) fa[np] = q;
        else{
            RG int nq = ++tot;
            len[nq] = len[p] + 1, fa[nq] = fa[q];
            for(RG int i = 0; i < 26; ++i) trans[i][nq] = trans[i][q];
            fa[q] = fa[np] = nq;
            while(p && trans[c][p] == q) trans[c][p] = nq, p = fa[p];
        }
    }
}

int main(RG int argc, RG char* argv[]){
    scanf(" %s", s), n = strlen(s);
    for(RG int i = 0; i < n; ++i) Extend(s[i] - 'a');
    scanf(" %s", s), n = strlen(s);
    for(RG int i = 0, nw = 1, cnt = 0; i < n; ++i){
        RG int c = s[i] - 'a';
        if(trans[c][nw]) ++cnt, nw = trans[c][nw];
        else{
            while(nw && !trans[c][nw]) nw = fa[nw], cnt = len[nw];
            if(!nw) nw = 1, cnt = 0;
            else cnt++, nw = trans[c][nw];
        }
        ans = max(ans, cnt);
    }
    printf("%d\n", ans);
    return 0;
}

转载于:https://www.cnblogs.com/cjoieryl/p/8900910.html