UVA 1297 最长回文串【裸】

题目大意：给一行字母，只有字母，最多1000个字母。 求最长回文穿，有多个情况下，输出最前面的那个。

这次终于让我后缀数组不TLE了…… 哇哇哇哇哇！  我就是喜欢用线段树写RMQ……慢就慢，我不管了

#include <cstdio>#include <iostream>#include <cstring>using namespace std;const int max_strlen = 4000  + 10;char text[max_strlen];int tub[max_strlen], wa[max_strlen], wb[max_strlen], wv[max_strlen];int R[max_strlen], height[max_strlen], rank[max_strlen];int totlen, SA[max_strlen];int textlen;bool 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)tub[i] = 0;for (i = 0; i != n; ++ i)++ tub[x[i] = r[i]];for (i = 1; i != m; ++ i)tub[i] += tub[i - 1];for (i = n - 1; i >= 0; -- i)sa[-- tub[x[i]]] = i;for (j = 1, p =1; p != n; m = p, j *= 2){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)tub[i] = 0;for (i = 0; i != n; ++ i)++ tub[wv[i]];for (i = 1; i != m; ++ i)tub[i] += tub[i - 1];for (i = n - 1; i >= 0; -- i)sa[-- tub[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], sa[i - 1], j) ? p - 1 : p ++;}}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 (j = sa[rank[i] - 1], k ? k -- : 0; r[i + k] == r[j + k]; ++ k);}void HZSZ(){totlen = 0;for (int i = 0; i != textlen; ++ i)R[totlen ++] = text[i];R[totlen ++] = 1; //串分割符号for (int i = textlen - 1; i >=0 ; -- i)R[totlen ++] = text[i];R[totlen] = 0; //结束符号da(R, SA, totlen + 1, 175);calheight(R, SA, totlen);}struct node{node *ls, *rs;int key, L, R;node(int LL, int RR, int KEY, node *LS, node *RS){ls = LS;rs = RS; L =LL;R = RR;key = KEY;}node(){ls = rs = this;key = L = R = -1;}}root, Tnull, *null = &Tnull;int mt(node &now, int LL, int RR){if (LL == RR)returnnow.key = height[LL];int mid = (LL + RR) / 2;now.ls = new node(LL, mid, 0, null, null);now.rs = new node(mid + 1, RR, 0, null, null);int a = mt(*now.ls, LL, mid);int b = mt(*now.rs, mid + 1, RR);return now.key = min(a, b);}int find(node &now, int LL, int RR){if (now.L == LL && now.R == RR)returnnow.key;int mid = (now.L + now.R) / 2;if (RR <= mid)return find(*now.ls, LL, RR);if (mid < LL)return find(*now.rs, LL, RR);int a = find(*now.ls, LL, mid);int b = find(*now.rs, mid + 1, RR);return min(a, b);}void doit(){root = node(0, totlen, 0, null, null);mt(root, 0, totlen);int ans = 1, flag = 1 ,wz = 0;for (int i = 1; i != textlen; ++ i){int a = rank[i];int b = rank[totlen - i - 1];if (a > b)swap(a, b);int tmp = (find(root, a + 1, b) * 2) - 1;if (tmp> ans){ans = tmp;;flag = 1;wz = i;}a = rank[i];b = rank[totlen - i];if (a > b)swap(a, b);tmp = find(root, a + 1, b) * 2;if (tmp > ans){ans = tmp;flag = 0;wz = i;}}if (flag == 1) //奇数的情况{int tmp = ans / 2;for (int i = wz - tmp; i < wz + tmp; ++ i)cout<<text[i];cout<<text[wz + tmp];}else{int tmp = ans / 2;for (int i = wz - tmp; i < wz + tmp - 1; ++ i)cout<<text[i];cout<<text[wz + tmp - 1];}}int main(){gets(text);textlen = strlen(text);HZSZ(); doit();return 0;}