POJ - 2774 Long Long Message (后缀数组)
题意：
求两个字符串的最长公共子串的长度。
思路：
利用后缀数组中 height[] 的意义，将两个字符串拼接为一个字符串，中间插入特殊符号；然后求得来自两个不同串的后缀的最长公共前缀即可。
1 /* 2 * @Author: WindyStreet 3 * @Date: 2018-07-31 14:30:43 4 * @Last Modified by: WindyStreet 5 * @Last Modified time: 2018-07-31 14:37:47 6 */ 7 #include<bits/stdc++.h> 8 9 using namespace std; 10 11 #define X first 12 #define Y second 13 #define eps 1e-2 14 #define gcd __gcd 15 #define pb push_back 16 #define PI acos(-1.0) 17 #define lowbit(x) (x)&(-x) 18 #define bug printf("!!!!!\n"); 19 #define mem(x,y) memset(x,y,sizeof(x)) 20 21 typedef long long LL; 22 typedef long double LD; 23 typedef pair<int,int> pii; 24 typedef unsigned long long uLL; 25 26 const int maxn = 2e5+7; 27 const int INF = 1<<30; 28 const int mod = 1e9+7; 29 char s[maxn],tmp[maxn]; 30 int sa[maxn],h[maxn],rk[maxn],c[maxn],t[maxn],t2[maxn]; 31 void work(int n,int m){ 32 int i, *x = t, *y = t2; 33 //基数排序 34 for(i = 0; i < m; i++) c[i] = 0; 35 for(i = 0; i < n; i++) c[x[i]=s[i]]++; 36 for(i = 1; i < m; i++) c[i] += c[i-1]; 37 for(i = n -1; i >= 0; i--) sa[--c[x[i]]] = i; 38 for(int k = 1; k <= n; k <<= 1){ 39 int p = 0 ; 40 for(i = n - k; i < n; i++) y[p++] = i; 41 for(i = 0; i < n; i++) if(sa[i] >= k) y[p++] = sa[i] - k; 42 for(i = 0; i < m; i++) c[i] = 0; 43 for(i = 0; i < n; i++) c[x[y[i]]]++; 44 for(i = 0; i < m; i++) c[i] += c[i-1]; 45 for(i = n-1; i >= 0; i--) sa[--c[x[y[i]]]] = y[i]; 46 swap(x,y); 47 p = 1; x[sa[0]] = 0; 48 for(i = 1; i < n ;i++){ 49 x[sa[i]] = y[sa[i-1]]==y[sa[i]]&&y[sa[i-1]+k]==y[sa[i]+k] ? p-1:p++; 50 } 51 if(p>=n)break; 52 m = p; 53 } 54 int k = 0; 55 for(i = 0; i < n; i++) rk[sa[i]] = i; 56 for(i = 0; i < n; i++){ 57 if(k) k--; 58 int j = sa[rk[i]-1]; 59 while(s[i+k]==s[j+k])k++; 60 h[rk[i]] = k; 61 } 62 } 63 64 void solve(){ 65 scanf("%s",s); 66 scanf("%s",tmp); 67 int len = strlen(s); 68 int len1= strlen(tmp); 69 s[len] = '#'; //中间插入特殊字符 70 for(int i = 0;i<len1;i++){ 71 s[len+i+1] = tmp[i]; 72 } 73 s[len + len1+1] = '\0'; 74 work(len+len1+1,128); 75 int ans = 0; 76 for(int i=2;i<=len+len1+1;i++){ 77 if(1LL*(sa[i] - len)*1ll*(sa[i-1] - len)<0) //判断是否来自两个不同的串 78 ans = max(ans, h[i]); 79 } 80 printf("%d\n",ans); 81 82 return; 83 } 84 85 int main() 86 { 87 // freopen("in.txt","r",stdin); 88 // freopen("out.txt","w",stdout); 89 // ios::sync_with_stdio(false); 90 int t = 1; 91 //scanf("%d",&t); 92 while(t--){ 93 // printf("Case %d: ",cas++); 94 solve(); 95 } 96 return 0; 97 }
