[hdu5340]二分，枚举，二进制压位加速

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......

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/* ******************************************************************************** */

#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;

}

/* ******************************************************************************** */

巴特西

[hdu5340]二分，枚举，二进制压位加速

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