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题意：判断一个字符串能否划成三段非空回文串。
思路：先用二分+hash在nlogn的时间内求出以每条对称轴为中心的回文串的最大半径r[i]（可以用对称的两个下标之和来表示 ），然后利用r[i]求出pre[i]和suf[i]，其中pre[i]表示0~i能否形成回文串，suf[i]表示i~n-1能否形成回文串。然后枚举中间的第二段回文串的对称轴，利用半径r[i]得到最左端L和最右端R，问题变成能否找到一个数d，使得pre[L+d] && suf[R-d]，-1<=d，L+d<R-d，这显然可以用二进制来加速，移位，按位与，判断是否为0等操作都降低了常数倍复杂度。理论上复杂度变为O(n^2/64)，实际上跑起来慢爆了，可能是写挫了TUT......
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 218 219 220 221 222 223 224 225 226 227 228 229 230 231 232 233 234 235 236 237 238 239 240 241 242 243 | /* ******************************************************************************** */#include <iostream> //#include <cstdio> //#include <cmath> //#include <cstdlib> //#include <cstring> //#include <vector> //#include <ctime> //#include <deque> //#include <queue> //#include <algorithm> //#include <map> //#include <cmath> //using namespace std; // //#define pb push_back //#define mp make_pair //#define X first //#define Y second //#define all(a) (a).begin(), (a).end() //#define fillchar(a, x) memset(a, x, sizeof(a)) // //typedef pair<int, int> pii; //typedef long long ll; //typedef unsigned long long ull; // //#ifndef ONLINE_JUDGE //void RI(vector<int>&a,int n){a.resize(n);for(int i=0;i<n;i++)scanf("%d",&a[i]);} //void RI(){}void RI(int&X){scanf("%d",&X);}template<typename...R> //void RI(int&f,R&...r){RI(f);RI(r...);}void RI(int*p,int*q){int d=p<q?1:-1; //while(p!=q){scanf("%d",p);p+=d;}}void print(){cout<<endl;}template<typename T> //void print(const T t){cout<<t<<endl;}template<typename F,typename...R> //void print(const F f,const R...r){cout<<f<<", ";print(r...);}template<typename T> //void print(T*p, T*q){int d=p<q?1:-1;while(p!=q){cout<<*p<<", ";p+=d;}cout<<endl;} //#endif // ONLINE_JUDGE //template<typename T>bool umax(T&a, const T&b){return b<=a?false:(a=b,true);} //template<typename T>bool umin(T&a, const T&b){return b>=a?false:(a=b,true);} //template<typename T> //void V2A(T a[],const vector<T>&b){for(int i=0;i<b.size();i++)a[i]=b[i];} //template<typename T> //void A2V(vector<T>&a,const T b[]){for(int i=0;i<a.size();i++)a[i]=b[i];} // //const double PI = acos(-1.0); //const int INF = 1e9 + 7; // ///* -------------------------------------------------------------------------------- */unsigned long long Pre[64], Suf[64];struct BitSet { vector<unsigned long long> s; static void init() { Pre[0] = 1; Suf[63] = (unsigned long long)1 << 63; for (int i = 1; i < 64; i ++) { Pre[i] = (unsigned long long)1 << i | Pre[i - 1]; } for (int i = 62; i >= 0; i --) { Suf[i] = (unsigned long long)1 << i | Suf[i + 1]; } } void resize(int n) { int p = s.size(), t = (n - 1) / 64 + 1; s.resize(t); } BitSet(int n) { resize(n); } BitSet() {} BitSet operator & (BitSet &that) { int sz = that.s.size(), n = this->s.size(), len = max(sz, n); if (sz < len) that.resize(len); if (n < len) this->resize(len); BitSet ans(len * 64); for (int i = len - 1; i >= 0; i --) { ans.s[i] = this->s[i] & that.s[i]; } return ans; } BitSet operator | (BitSet &that) { int sz = that.s.size(), n = this->s.size(), len = max(sz, n); if (sz < len) that.resize(len); if (n < len) this->resize(len); BitSet ans(len * 64); for (int i = len - 1; i >= 0; i --) { ans.s[i] = this->s[i] | that.s[i]; } return ans; } BitSet operator ^ (BitSet &that) { int sz = that.s.size(), n = this->s.size(), len = max(sz, n); if (sz < len) that.resize(len); if (n < len) this->resize(len); BitSet ans(len * 64); for (int i = len - 1; i >= 0; i --) { ans.s[i] = this->s[i] ^ that.s[i]; } return ans; } BitSet operator << (int x) { int sz = s.size(), c = x / 64, r = x % 64; BitSet ans(64 * sz); for (int i = sz - 1; i - c >= 0; i --) { ans.s[i] = (s[i - c] & Pre[63 - r]) << r; if (r && i - c - 1 >= 0) ans.s[i] |= (s[i - c - 1 ] & Suf[64 - r]) >> (64 - r); } return ans; } BitSet operator >> (int x) { int sz = s.size(), c = x / 64, r = x % 64; BitSet ans(64 * sz); for (int i = 0; i + c < sz; i ++) { ans.s[i] = (s[i + c] & Suf[r]) >> r; if (r && i + c + 1 < sz) ans.s[i] |= (s[i + c + 1] & Pre[r - 1]) << (64 - r); } return ans; } bool get(int p) { int c = p / 64, r = p % 64; return s[c] & ((unsigned long long)1 << r); } bool zero() { int n = s.size(); for (int i = 0; i < n; i ++) { if (s[i]) return false; } return true; } void setval(int L, int R, bool val) { int p = L / 64, tp = L % 64, q = R / 64, tq = R % 64; for (int i = p + 1; i < q; i ++) { s[i] = val? ((unsigned long long)-1) : 0; } if (p == q) { unsigned long long buf = Suf[tp] & Pre[tq]; s[p] = val? s[p] | buf : s[p] & ~buf; return ; } s[p] = val? s[p] | Suf[tp] : s[p] & ~Suf[tp]; s[q] = val? s[q] | Pre[tq] : s[q] & ~Pre[tq]; } void print() { int n = s.size(); for (int i = n - 1; i >= 0; i --) { unsigned long long x = s[i]; for (int i = 63; i >= 0; i --) { if (((unsigned long long)1 << i) & x) putchar(‘1‘); else putchar(‘0‘); } } putchar(‘\n‘); }};struct StringHash { const static unsigned int hack = 1301; const static int maxn = 1e5 + 7; unsigned long long H[maxn], C[maxn]; void init(char s[], int n) { for (int i = 0; s[i]; i ++) { H[i] = (i? H[i - 1] * hack : 0) + s[i]; } C[0] = 1; for (int i = 1; i <= n; i ++) C[i] = C[i - 1] * hack; } unsigned long long get(int L, int R) { return H[R] - (L? H[L - 1] * C[R - L + 1] : 0); }} ;StringHash hsh, hshrev;const int maxn = 1e5 + 7;bool pre[maxn], suf[maxn];char s[maxn], revs[maxn];int F[maxn];int main() {#ifndef ONLINE_JUDGE freopen("in.txt", "r", stdin);#endif // ONLINE_JUDGE BitSet::init(); int T; cin >> T; while (T --) { scanf("%s", s); int n = strlen(s), total = n * 2 - 1; hsh.init(s, n); for (int i = 0; i < n; i ++) revs[i] = s[n - i - 1]; hshrev.init(revs, n); for (int i = 0; i < total; i ++) { int L = i / 2, R = (i + 1) / 2; int minlen = 0, maxlen = min(L + 1, n - R); while (minlen < maxlen) { int midlen = (minlen + maxlen + 1) >> 1; int lpos = L - midlen + 1, rpos = R + midlen - 1; if (hsh.get(lpos, L) == hshrev.get(n - rpos - 1, n - R - 1)) minlen = midlen; else maxlen = midlen - 1; } F[i] = minlen; } fillchar(pre, 0); fillchar(suf, 0); pre[0] = suf[n - 1] = true; BitSet bs1(n), bs2(n); bs1.setval(0, 0, 1); bs2.setval(0, 0, 1); for (int i = 1; i < n; i ++) { pre[i] = F[i] == i / 2 + 1; if (pre[i]) bs1.setval(i, i, 1); } for (int i = n - 2; i >= 0; i --) { suf[i] = F[i + n - 1] == (n - i + 1) / 2; if (suf[i]) bs2.setval(n - i - 1, n - i - 1, 1); } bool ok = false; for (int i = 0; i < total; i ++) { int L = i / 2, R = (i + 1) / 2; int len = F[i], lpos = L - len + 1, rpos = R + len - 1; if (len == 0) continue; BitSet buf, result, newbuf(n); if (n - R >= L + 1) { buf = bs2 >> (n - R - L - 1); result = bs1 & buf; newbuf.setval(max(0, lpos - 1), L - 1, 1); } if (L + 1 > n - R) { buf = bs1 >> (L + 1 - n + R); result = buf & bs2; newbuf.setval(max(0, n - rpos - 2), n - R - 2, 1); } newbuf = newbuf & result; if (!newbuf.zero()) { ok = true; break; } } puts(ok? "Yes" : "No"); } return 0;}/* ******************************************************************************** */ |
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原文地址：http://www.cnblogs.com/jklongint/p/4695077.html
