spoj 1812 Longest Common Substring II（后缀自动机）

最新推荐文章于 2020-01-27 10:07:07 发布

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本文介绍了一个解决 SPOJ 问题 1812 (Longest Common Substring II) 的算法实现。通过使用后缀自动机 (SAM) 方法来高效地找到两个字符串之间的最长公共子串。该解决方案首先建立后缀自动机，然后进行匹配过程以更新最长公共子串长度。

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题目链接：spoj 1812 Longest Common Substring II

代码

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
typedef long long ll;
const int maxn = 100005;
const int SIGMA_SIZE = 26;
const int inf = 0x3f3f3f3f;

struct SAM {
    int sz, last;
    int g[maxn<<1][SIGMA_SIZE], pre[maxn<<1], step[maxn<<1];
    int lcs[maxn<<1], tlcs[maxn<<1];

    void newNode(int s) {
        step[++sz] = s;
        pre[sz] = 0;
        lcs[sz] = s;
        memset(g[sz], 0, sizeof(g[sz]));
    }

    int idx(char ch) { return ch -'a'; }

    void init() {
        sz = 0, last = 1;
        newNode(0);
        memset(tlcs, 0, sizeof(tlcs));
    }

    void insert(char ch) {
        newNode(step[last] + 1);
        int v = idx(ch), p = last, np = sz;

        while (p && !g[p][v]) {
            g[p][v] = np;
            p = pre[p];
        }

        if (p) {
            int q = g[p][v];
            if (step[q] == step[p] + 1)
                pre[np] = q;
            else {
                newNode(step[p] + 1);
                int nq = sz;
                for (int j = 0; j < SIGMA_SIZE; j++) g[nq][j] = g[q][j];

                pre[nq] = pre[q];
                pre[np] = pre[q] = nq;

                while (p && g[p][v] == q) {
                    g[p][v] = nq;
                    p = pre[p];
                }
            }
        } else
            pre[np] = 1;
        last = np;
    }

    void match(char* str) {
        memset(tlcs, 0 , sizeof(tlcs));
        int n = strlen(str), p = 1, ans = 0;
        for (int i = 0; i < n; i++) {
            int v = idx(str[i]);
            if (g[p][v]) p = g[p][v], ans++;
            else {
                while (p && g[p][v] == 0) p = pre[p];
                if (p) ans = step[p] + 1, p = g[p][v];
                else ans = 0, p = 1;
            }

            tlcs[p] = max(tlcs[p], ans);
        }
        for (int i = 0; i < sz; i++) {
            int u = id[i];
            lcs[u] = min(lcs[u], tlcs[u]);
            int v = pre[u];
            //printf("%d %d %d!", u, v, tlcs[u]);
            tlcs[v] = max(tlcs[v], tlcs[u]);
        }
        //printf("\n");
    }

    int solve() {
        int ans = 0;
        for (int i = 1; i <= sz; i++) {
            //printf("%d ", lcs[i]);
            ans = max(ans, lcs[i]);
        }
        //printf("\n");
        return ans;
    }

    int du[maxn<<1], id[maxn<<1];
    void topu() {
        memset(du, 0, sizeof(du));
        for (int i = 1; i <= sz; i++)
            du[pre[i]]++;

        int head = 0, rear = 0;
        for (int i = 1; i <= sz; i++)
            if (du[i] == 0) id[rear++] = i;

        while (head < rear) {
            int u = id[head++];
            du[pre[u]]--;
            if (du[pre[u]] == 0 && pre[u])
                id[rear++] = pre[u];
        }
    }
}SA;

char str[20][maxn];

int main () {
    int N = 0;
    while (gets(str[N]) != NULL) N++;

    SA.init();
    int len = strlen(str[0]);
    for (int i = 0; i < len; i++) SA.insert(str[0][i]);

    SA.topu();
    for (int i = 1; i < N; i++) SA.match(str[i]);
    printf("%d\n", SA.solve());
    return 0;
}