HDU - 5371 Hotaru's problem 线段树 + hash

HDU - 5371

求出每个点的最大回文长度。

然后转换成线段互相覆盖的最大长度。

然后离线之后用线段树 或者 set维护一下就好了。

#include<bits/stdc++.h>
#define LL long long
#define LD long double
#define ull unsigned long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ALL(x) (x).begin(), (x).end()
#define fio ios::sync_with_stdio(false); cin.tie(0);

using namespace std;

const int N = 2e5 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-8;
const double PI = acos(-1);

template<class T, class S> inline void add(T &a, S b) {a += b; if(a >= mod) a -= mod;}
template<class T, class S> inline void sub(T &a, S b) {a -= b; if(a < 0) a += mod;}
template<class T, class S> inline bool chkmax(T &a, S b) {return a < b ? a = b, true : false;}
template<class T, class S> inline bool chkmin(T &a, S b) {return a > b ? a = b, true : false;}

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

int n, a[N], b[N];
int maxLen[N];

struct StrHash {
    const ull B = 23333;
    ull hsL[N], hsR[N], Pow[N];
    void build(int *a, int n) {
        Pow[0] = 1;
        hsR[n + 1] = 0;
        for(int i = 1; i <= n; i++) {
            hsL[i] = hsL[i - 1] * B + a[i];
            Pow[i] = Pow[i - 1] * B;
        }
        for(int i = n; i >= 1; i--) {
            hsR[i] = hsR[i + 1] * B + a[i];
        }
    }
    ull getLVal(int L, int R) {
        return hsL[R] - hsL[L - 1] * Pow[R - L + 1];
    }
    ull getRVal(int L, int R) {
        return hsR[L] - hsR[R + 1] * Pow[R - L + 1];
    }
} ha;

struct SegmentTree {
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
    int mx[N << 2];
    void build(int l, int r, int rt) {
        mx[rt] = 0;
        if(l == r) return;
        int mid = l + r >> 1;
        build(lson); build(rson);
    }
    void update(int p, int val, int l, int r, int rt) {
        if(l == r) {
            mx[rt] = val;
            return;
        }
        int mid = l + r >> 1;
        if(p <= mid) update(p, val, lson);
        else update(p, val, rson);
        mx[rt] = max(mx[rt << 1], mx[rt << 1 | 1]);
    }
    int query(int R, int l, int r, int rt) {
        if(mx[rt] < R) return -1;
        if(l >= R) return -1;
        if(l == r) return l;
        int mid = l + r >> 1;
        if(mx[rt << 1] >= R) return query(R, lson);
        else return query(R, rson);
    }
} Tree;

struct Line {
    int l, r;
    bool operator < (const Line &rhs) const {
        return l < rhs.l;
    }
} L[N];

int main() {

    int T; scanf("%d", &T);
    for(int cas = 1; cas <= T; cas++) {
        scanf("%d", &n);
        for(int i = 1; i <= n; i++) {
            scanf("%d", &a[i]);
        }
        for(int i = 1; i <= n; i++) {
            b[i << 1] = a[i];
        }
        n = 2 * n + 1;
        for(int i = 1; i <= n; i += 2) {
            b[i] = 100000001;
        }
        ha.build(b, n);
        Tree.build(1, n, 1);
        int Ltot = 0;
        for(int i = 1; i <= n; i += 2) {
            int low = 1, high = min(i, n - i + 1);
            while(low <= high) {
                int mid = low + high >> 1;
                if(ha.getLVal(i - mid + 1, i) == ha.getRVal(i, i + mid - 1)) {
                    maxLen[i] = mid;
                    low = mid + 1;
                }
                else {
                    high = mid - 1;
                }
            }
            L[++Ltot] = Line{i - maxLen[i] + 1, i};
        }
        for(int i = 1; i <= n; i += 2) {
            Tree.update(i, i + maxLen[i] - 1, 1, n, 1);
        }
        int ans = 0;
        sort(L + 1, L + 1 + Ltot);
        for(int i = 1, j = 1; i <= Ltot; i++) {
            int l = L[i].l, r = L[i].r;
            if(l == r) continue;
            while(j <= n && j < l) {
                Tree.update(j, 0, 1, n, 1);
                j += 2;
            }
            int p = Tree.query(r, 1, n, 1);
            if(~p) {
                chkmax(ans, (r - p) >> 1);
            }
        }
        printf("Case #%d: %d\n", cas, ans * 3);
    }
    return 0;
}

/*
*/