挑战程序竞赛系列（71）：4.7高度数组（1）

挑战程序竞赛系列（71）：4.7高度数组（1）

传送门：POJ 2217: Secretary

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题意：

给定两个字符串S和T。请计算两个字符串最长的公共字符串子串的长度。

LCP概念：在后缀数组中相邻两个后缀的最长公共前缀（Longest Common Prefix）

的确很取巧，相邻位置的最长公共前缀可以由前一个位置的最长公共前缀得出，至少在h-1个位置上不需要再比较，而直接比较后续的字符即可。

代码如下：

import java.io.IOException;import java.util.Arrays;import java.util.Comparator;import java.util.Scanner;public class Main{    String INPUT = "./data/judge/201709/P2217.txt";    static Scanner in;    public static void main(String[] args) throws IOException {        in = new Scanner(System.in);        new Main().run();    }    void read() {        int t = in.nextInt();        in.nextLine();        while (t --> 0) {            String S = in.nextLine();            String T = in.nextLine();            StringBuilder sb = new StringBuilder();            sb.append(S + '$' + T);            SuffixArray sa = new SuffixArray(sb.toString().toCharArray());            int max = 0;            for (int i = 0; i < sb.length(); ++i) {                if (sa.sa[i] < S.length() != sa.sa[i + 1] < S.length())                    max = Math.max(max, sa.lcp[i]);            }            System.out.println("Nejdelsi spolecny retezec ma delku " + max + ".");        }    }    class SuffixArray{        int k = 1;        int n;        Integer[] sa;        int[] rank, tmp;        int[] lcp;        SuffixArray(char[] cs){            n    = cs.length;            sa   = new Integer[n + 1];            rank = new int[n + 1];            tmp  = new int[n + 1];            lcp  = new int[n];            for (int i = 0; i <= n; ++i) {                sa[i] = i;                rank[i] = i < n ? cs[i] : -1;            }            for (k = 1; k <= n; k <<= 1) {                Arrays.sort(sa, cmp);                tmp[sa[0]] = 0;                for (int i = 1; i <= n; ++i) {                    tmp[sa[i]] = tmp[sa[i - 1]] + (cmp.compare(sa[i - 1], sa[i]) < 0 ? 1 : 0);                  }                for (int i = 0; i <= n; ++i) {                    rank[i] = tmp[i];                }            }            // lcp            for (int i = 0; i <= n; ++i) rank[sa[i]] = i;  //求出每个后缀的当前排名            int h = 0;            for (int i = 0; i < n; ++i) {                int j = sa[rank[i] - 1]; // 后缀i的前一个后缀                if (h > 0) h--;                for (;j + h < n && i + h < n; ++h) {                    if (cs[j + h] != cs[i + h]) break;                }                lcp[rank[i] -1] = h;            }        }        private Comparator<Integer> cmp = new Comparator<Integer>() {            @Override            public int compare(Integer o1, Integer o2) {                int i = o1;                int j = o2;                if (rank[i] != rank[j]) return rank[i] - rank[j];                else {                    int ri = i + k <= n ? rank[i + k] : -1;                    int rj = j + k <= n ? rank[j + k] : -1;                    return ri - rj;                }            }        };    }    void run() throws IOException {        read();    }}