题目大意:
给出n( n <= 100)个字符串, 只包含小写字母, 每个字符串长度不超过1000且不为空
求出最长的子串,满足在n个字符串当中出现在一半以上的字符串上, 如果有多个这样的子串,按字典序输出
大致思路:
简单的后缀数组height数组分组的运用, 首先将所有的串都连接起来, 中间用不同的没有出现在n个字符串中的字符隔开, 然后二分子串长度L判断是否存在满足条件的长度为子串, 对于多个解用vector存储一下其起始位置,最后还原字符串排序后输出即可
一发AC (~。~)
代码如下:
Result : Accepted Memory : 3776 KB Time : 532 ms
/* * Author: Gatevin * Created Time: 2015/2/11 12:52:18 * File Name: Mononobe_Mitsuki.cpp */ #include<iostream> #include<sstream> #include<fstream> #include<vector> #include<list> #include<deque> #include<queue> #include<stack> #include<map> #include<set> #include<bitset> #include<algorithm> #include<cstdio> #include<cstdlib> #include<cstring> #include<cctype> #include<cmath> #include<ctime> #include<iomanip> using namespace std; const double eps(1e-8); typedef long long lint; #define maxn 102333 /* * 先将所有串连接起来,中间用不同的没有出现的值隔开,求出后缀数组和height数组后 * 利用height数组进行分组, 二分判断是否存在长度长度为L的串满足条件 * 将满足条件的长度对应的多个不同子串的开始位置保存起来即可, 简单的height数组分组应用 */ /* * Doubling Algorithm求后缀数组 */ int wa[maxn], wb[maxn], wv[maxn], Ws[maxn]; int cmp(int *r, int a, int b, int l) { return r[a] == r[b] && r[a + l] == r[b + l]; } void da(int *r, int *sa, int n, int m) { int *x = wa, *y = wb, *t, i, j, p; for(i = 0; i < m; i++) Ws[i] = 0; for(i = 0; i < n; i++) Ws[x[i] = r[i]]++; for(i = 1; i < m; i++) Ws[i] += Ws[i - 1]; for(i = n - 1; i >= 0; i--) sa[--Ws[x[i]]] = i; for(j = 1, p = 1; p < n; j *= 2, m = p) { for(p = 0, i = n - j; i < n; i++) y[p++] = i; for(i = 0; i < n; i++) if(sa[i] >= j) y[p++] = sa[i] - j; for(i = 0; i < n; i++) wv[i] = x[y[i]]; for(i = 0; i < m; i++) Ws[i] = 0; for(i = 0; i < n; i++) Ws[wv[i]]++; for(i = 1; i < m; i++) Ws[i] += Ws[i - 1]; for(i = n - 1; i >= 0; i--) sa[--Ws[wv[i]]] = y[i]; for(t = x, x = y, y = t, p = 1, x[sa[0]] = 0, i = 1; i < n; i++) x[sa[i]] = cmp(y, sa[i - 1], sa[i], j) ? p - 1 : p++; } return; } int rank[maxn], height[maxn]; void calheight(int *r, int *sa, int n) { int i, j, k = 0; for(i = 1; i <= n; i++) rank[sa[i]] = i; for(i = 0; i < n; height[rank[i++]] = k) for(k ? k-- : 0, j = sa[rank[i] - 1]; r[i + k] == r[j + k]; k++); return; } char in[1010]; int s[maxn], sa[maxn], belong[maxn]; int n; vector <int> pos[1010];//pos[i]保存长度为i的满足条件的串的起始位置,相同的只记录一次 bool check(int mid, int N)//查找是否存在满足条件的长度为mid的串 { bool vis[110]; memset(vis, 0, sizeof(vis)); int cnt = 0; bool ret = false, in = false; for(int i = 1; i <= N; i++) if(height[i] >= mid) { if(!vis[belong[sa[i]]]) vis[belong[sa[i]]] = 1, cnt++; if(cnt > (n >> 1)) { ret = true; if(!in) { in = true; pos[mid].push_back(sa[i]); } } } else memset(vis, 0, sizeof(vis)), vis[belong[sa[i]]] = 1, cnt = 1, in = false; return ret; } int main() { int N; bool between = false; while(scanf("%d", &n), n) { N = 0; if(between) printf("\n"); else between = true; for(int i = 0; i <= 1000; i++) pos[i].clear(); for(int i = 0; i < n; i++) { scanf("%s", in); int tmp = strlen(in); for(int j = 0; j < tmp; j++) belong[N] = i, s[N++] = in[j] - 'a' + 1; s[N++] = 27 + i; } N--; s[N] = 0; da(s, sa, N + 1, 130); calheight(s, sa, N); int L = 0, R = 1000, ans = 0, mid; while(L <= R)//二分查找最大的长度mid { mid = (L + R) >> 1; if(check(mid, N)) { ans = mid; L = mid + 1; } else R = mid - 1; } if(ans == 0) printf("?\n"); else //由后缀数组的性质本身已经是字典序升序了 for(unsigned int i = 0; i < pos[ans].size(); i++) { for(int j = 0; j < ans; j++) printf("%c", s[pos[ans][i] + j] - 1 + 'a'); printf("\n"); } } return 0; }