hdu1403 Longest Common Substring

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原文链接：http://www.cnblogs.com/weeping/p/6648830.html

文章标签：

#java #php

本文介绍了一种求解两个字符串最长公共子串的有效算法。通过将两个字符串连接并使用特殊字符分隔，利用二分查找及高度分组的方法确定最长公共子串的长度。文章提供了一个完整的C++实现示例。

地址：http://acm.split.hdu.edu.cn/showproblem.php?pid=1403

题目：

Longest Common Substring

Time Limit: 8000/4000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6296 Accepted Submission(s): 2249

Problem Description

Given two strings, you have to tell the length of the Longest Common Substring of them.

For example:
str1 = banana
str2 = cianaic

So the Longest Common Substring is "ana", and the length is 3.

Input

The input contains several test cases. Each test case contains two strings, each string will have at most 100000 characters. All the characters are in lower-case.

Process to the end of file.

Output

For each test case, you have to tell the length of the Longest Common Substring of them.

Sample Input

banana cianaic

Sample Output

Author

Ignatius.L

思路：把两个字符串连接起来，中间用一个没出现过的字符隔开。

　　然后二分答案，二分check时对height进行分组，判断height值全大于x的组内是否同时包含两个字符串的子串

 1 #include <cstdlib>
 2 #include <cstring>
 3 #include <cstdio>
 4 #include <algorithm>
 5 
 6 const int N = 200005;
 7 int sa[N],s[N],wa[N], wb[N], ws[N], wv[N];
 8 int rank[N], height[N];
 9 
10 bool cmp(int r[], int a, int b, int l)
11 {
12     return r[a] == r[b] && r[a+l] == r[b+l];
13 }
14 
15 void da(int r[], int sa[], int n, int m)
16 {
17     int i, j, p, *x = wa, *y = wb;
18     for (i = 0; i < m; ++i) ws[i] = 0;
19     for (i = 0; i < n; ++i) ws[x[i]=r[i]]++;
20     for (i = 1; i < m; ++i) ws[i] += ws[i-1];
21     for (i = n-1; i >= 0; --i) sa[--ws[x[i]]] = i;
22     for (j = 1, p = 1; p < n; j *= 2, m = p)
23     {
24         for (p = 0, i = n - j; i < n; ++i) y[p++] = i;
25         for (i = 0; i < n; ++i) if (sa[i] >= j) y[p++] = sa[i] - j;
26         for (i = 0; i < n; ++i) wv[i] = x[y[i]];
27         for (i = 0; i < m; ++i) ws[i] = 0;
28         for (i = 0; i < n; ++i) ws[wv[i]]++;
29         for (i = 1; i < m; ++i) ws[i] += ws[i-1];
30         for (i = n-1; i >= 0; --i) sa[--ws[wv[i]]] = y[i];
31         for (std::swap(x, y), p = 1, x[sa[0]] = 0, i = 1; i < n; ++i)
32             x[sa[i]] = cmp(y, sa[i-1], sa[i], j) ? p-1 : p++;
33     }
34 }
35 
36 void calheight(int r[], int sa[], int n)
37 {
38     int i, j, k = 0;
39     for (i = 1; i <= n; ++i) rank[sa[i]] = i;
40     for (i = 0; i < n; height[rank[i++]] = k)
41         for (k?k--:0, j = sa[rank[i]-1]; r[i+k] == r[j+k]; k++);
42 }
43 bool check(int la,int lb,int lc,int x)
44 {
45     int m1=0,m2=0;
46     if(sa[1]<la)m1=1;
47     if(sa[1]>la)m2=1;
48     for(int i=2;i<=lc;i++)
49     {
50         if(height[i]<x)
51         {
52             if(m1&&m2)
53                 return 1;
54             m1=m2=0;
55         }
56         if(sa[i]<la)m1=1;
57         if(sa[i]>la)m2=1;
58     }
59     return m1&&m2;
60 }
61 char ss[N];
62 int main()
63 {
64     while(scanf("%s",ss)==1)
65     {
66         int la=strlen(ss),lb,n=0;
67         for(int i=0;i<la;i++)
68             s[n++]=ss[i]-'a'+1;
69         s[n++]=28;
70         scanf("%s",ss);
71         lb=strlen(ss);
72         for(int i=0;i<lb;i++)
73             s[n++]=ss[i]-'a'+1;
74         s[n]=0;
75         da(s,sa,n+1,30);
76         calheight(s,sa,n);
77         int l=1,r=la,ans=0;
78         while(l<=r)
79         {
80             int mid=l+r>>1;
81             if(check(la,lb,n,mid))
82                 ans=mid,l=mid+1;
83             else
84                 r=mid-1;
85         }
86         printf("%d\n",ans);
87     }
88     return 0;
89 }