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.
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.
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
3
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
",ans);
87 }
88 return 0;
89 }