POJ2774 Long Long Message
两串求最长公共子串
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 using namespace std; 6 const int maxn = 2e5+5; 7 char s[maxn]; 8 int sa[maxn], t[maxn], t2[maxn], c[maxn]; 9 int n; 10 //构造字符串s的后缀数组， 每个字符值必须为0 ～ m-1 11 void build_sa(int m) { 12 int *x = t, *y = t2; 13 //基数排序 14 for(int i = 0; i < m; i++) c[i] = 0; 15 for(int i = 0; i < n; i++) c[x[i] = s[i]]++; 16 for(int i = 1; i < m; i++) c[i] += c[i-1]; 17 for(int i = n-1; i >= 0; i--) sa[--c[x[i]]] = i; 18 for(int k = 1; k <= n; k <<= 1) { 19 int p = 0; 20 //直接利用sa数组排序第二关键字 21 for(int i = n-k; i < n; i++) y[p++] = i; 22 for(int i = 0; i < n; i++) if(sa[i] >= k) y[p++] = sa[i] - k; 23 //基数排序第一关键字 24 for(int i = 0; i < m; i++) c[i] = 0; 25 for(int i = 0; i < n; i++) c[x[y[i]]]++; 26 for(int i = 1; i < m; i++) c[i] += c[i-1]; 27 for(int i = n-1; i>= 0; i--) sa[--c[x[y[i]]]] = y[i]; 28 //根据sa和y数组计算新的x数组 29 swap(x, y); 30 p = 1; 31 x[sa[0]] = 0; 32 for(int i = 1; i < n; i++) 33 x[sa[i]] = (y[sa[i-1]] == y[sa[i]] && y[sa[i-1]+k] == y[sa[i]+k] ? p-1 : p++); 34 if(p >= n) break; 35 m = p; 36 } 37 } 38 39 int rank_[maxn]; //rank[i]代表后缀i在sa数组中的下标 40 int height[maxn]; //height[i] 定义为sa[i-1] 和 sa[i] 的最长公共前缀 41 //后缀j和k的LCP长度等于RMQ(height, rank[j]+1, rank[k]) 42 void get_height() { 43 int i, j, k = 0; 44 for(int i = 0; i < n; i++) rank_[sa[i]] = i; 45 for(int i = 0; i < n; i++) { 46 if(!rank_[i]) continue; 47 int j = sa[rank_[i]-1]; 48 if(k) k--; 49 50 while(s[i+k] == s[j+k]) k++; 51 height[rank_[i]] = k; 52 } 53 } 54 55 char s1[maxn], s2[maxn]; 56 int main() { 57 scanf("%s %s",s1,s2); 58 int len1 = strlen(s1); 59 int len2 = strlen(s2); 60 61 int cnt = 0; 62 for (int i = 0; i < len1; i++) 63 s[cnt++] = s1[i]; 64 s[cnt++] = 1; 65 for (int i = 0; i < len2; i++) 66 s[cnt++] = s2[i]; 67 s[cnt] = '\0'; 68 n = strlen(s) + 1; 69 build_sa(128); 70 get_height(); 71 72 int ans = 0; 73 int be; 74 for (int i = 2; i < cnt; i++) { 75 if (sa[i-1] < len1 && sa[i] > len1) { 76 if (ans < height[i]) { 77 be = sa[i-1]; 78 ans = height[i]; 79 } 80 } 81 else if (sa[i-1] > len1 && sa[i] < len1) { 82 if (ans < height[i]) { 83 be = sa[i]; 84 ans = height[i]; 85 } 86 } 87 } 88 printf("%d\n",ans); 89 /* 90 printf("pos: %d\n", be); 91 for (int i = 0; i < ans; i++) 92 putchar(s[be+i]); 93 putchar('\n'); 94 */ 95 return 0; 96 }
