标签:blog io os ar for sp div 问题 on
本题是利用后缀数组求最长的回文串。
方法是将字符串反转之后拼接到原来的字符串末尾,中间用一个没有出现过的分割符隔开,原因是防止最长公共前缀横跨两个串。
之后分别枚举回文串的中点,以及回文串长度是奇数还是偶数,看一下对应位置的最长公共前缀即可。
这里的求最长公共前缀要处理RMQ问题,线段树固然可以解决,但是显然ST 算法更加快一些。
#include <cstdio> #include <cstring> #include <algorithm> #include <queue> #include <stack> #include <map> #include <set> #include <climits> #include <string> using namespace std; #define MP make_pair #define PB push_back typedef long long LL; typedef unsigned long long ULL; typedef vector<int> VI; typedef pair<int, int> PII; typedef pair<double, double> PDD; const int INF = INT_MAX / 3; const double eps = 1e-8; const LL LINF = 1e17; const double DINF = 1e60; const int maxn = 5005; //以下是倍增法求后缀数组 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 i, j, p, *x = wa, *y = wb, *t; 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 <<= 1, 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 = 0; 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++; } } //以下是求解height 数组 int height[maxn], Rank[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++) ; } char buf[maxn]; int str[maxn], sa[maxn], len, n; int minv[maxn][20]; void init_RMQ() { for(int i = 0; i <= len; i++) minv[i][0] = height[i]; for(int j = 1; (1 << j) <= len + 1; j++) { for(int i = 0; i + (1 << j) - 1 <= len; i++) { minv[i][j] = min(minv[i][j - 1], minv[i + (1 << (j - 1))][j - 1]); } } } int query(int ql, int qr) { if(ql > qr) swap(ql, qr); ql++; int k = 0; while((1 << (k + 1)) <= qr - ql + 1) k++; return min(minv[ql][k], minv[qr - (1 << k) + 1][k]); } int main() { while(scanf("%s", buf) != EOF) { n = len = strlen(buf); for(int i = 0; i < len; i++) str[i] = buf[i] + 1; str[len++] = ‘*‘ + 1; for(int i = n - 1; i >= 0; i--) str[len++] = buf[i] + 1; str[len] = 0; da(str, sa, len + 1, 200); calheight(str, sa, len); init_RMQ(); int anslen = 1, anspos = 0; for(int i = 0; i < n; i++) { int nowval = query(Rank[i], Rank[len - i - 1]); int nowlen = nowval * 2 - 1; if(nowlen > anslen) { anslen = nowlen; anspos = i - nowval + 1; } if(i == 0) continue; nowval = query(Rank[i], Rank[len - i]); nowlen = nowval * 2; if(nowlen > anslen) { anslen = nowlen; anspos = i - nowval; } } for(int i = 0; i < anslen; i++) putchar(buf[i + anspos]); puts(""); } return 0; }
标签:blog io os ar for sp div 问题 on
原文地址:http://www.cnblogs.com/rolight/p/3995309.html