[hdu5371 Hotaru's problem]最大回文半径

题意：在一个字符串里面找最长的[A][B][A]子串，其中[A][B]是回文串，[A]和[B]的长度相等

思路：[A][B]是回文串，所以[B][A]也是回文串。先预处理出每个点的最大回文半径Ri，枚举[A][B]的对称轴位置p，那么就是要找最大的一个[B][A]的对称轴位置i，满足i<=p+R[p]，i-R[i]<=p。由于p是递增的，先前满足的以后肯定满足，于是可以用set来维护i-R[i]<=p的所有的位置i的集合，并可在logN的时间内得到最大的位置i。

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#include <map> #include <set> #include <cmath> #include <ctime> #include <deque> #include <queue> #include <stack> #include <vector> #include <cstdio> #include <string> #include <cstdlib> #include <cstring> #include <iostream> #include <algorithm> using namespace std; #define X first #define Y second #define pb push_back #define mp make_pair #define all(a) (a).begin(), (a).end() #define fillchar(a, x) memset(a, x, sizeof(a)) #define copy(a, b) memcpy(a, b, sizeof(a)) typedef long long ll; typedef pair<int, int> pii; 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 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; const double EPS = 1e-8; /* -------------------------------------------------------------------------------- */ const int maxn = 2e5 + 7; struct StringHash { const static unsigned int hack = 79; //const static int maxn = 2e5 + 7; unsigned long long H[maxn], C[maxn]; void init(int s[], int n) { for (int i = 0; i < n; 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; vector<int> G[maxn]; int a[maxn], b[maxn], F[maxn]; set<int> S; int main() { #ifndef ONLINE_JUDGE freopen("in.txt", "r", stdin); //freopen("out.txt", "w", stdout); #endif // ONLINE_JUDGE int T, n, cas = 0; cin >> T; while (T --) { cin >> n; for (int i = 0; i < n; i ++) { scanf("%d", a + i); } hsh.init(a, n); for (int i = 0; i < n; i ++) b[i] = a[n - i - 1]; hshrev.init(b, n); int total = 2 * n - 1; 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; } S.clear(); for (int i = 0; i < n; i ++) G[i].clear(); int ans = 0; for (int i = 0; i < n - 1; i ++) { G[i - F[2 * i + 1] + 1].pb(i); } for (int i = 0; i < G[0].size(); i ++) S.insert(G[0][i]); for (int i = 1; i < total; i += 2) { int L = i / 2; for (int j = 0; j < G[L + 1].size(); j ++) S.insert(G[L + 1][j]); if (S.size()) { set<int>::iterator R = S.upper_bound(L + F[i]); R --; umax(ans, *R - L); } } printf("Case #%d: %d ", ++ cas, ans * 3); } return 0; }