[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。

 

  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
 
#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\n", ++ cas, ans * 3);
    }

    return 0;
}
posted @ 2015-08-12 13:41  jklongint  阅读(263)  评论(0)  编辑  收藏  举报