• 【LOJ】#2073. 「JSOI2016」扭动的回文串


    题解

    就是一个回文串拼上左右两端
    类似二分找lcp这么做
    可以直接用哈希找回文串

    注意要找A串前半部分,B串找后半部分

    代码

    #include <bits/stdc++.h>
    #define enter putchar('
    ')
    #define space putchar(' ')
    #define pii pair<int,int>
    #define fi first
    #define se second
    #define mp make_pair
    #define MAXN 100005
    #define mo 99994711
    #define pb push_back
    //#define ivorysi
    using namespace std;
    typedef long long int64;
    typedef double db;
    template<class T>
    void read(T &res) {
        res = 0;T f = 1;char c = getchar();
        while(c < '0' || c > '9') {
            if(c == '-') f = -1;
            c = getchar();
        }
        while(c >= '0' && c <= '9') {
            res = res * 10 + c - '0';
            c = getchar();
        }
        res *= f;
    }
    template<class T>
    void out(T x) {
        if(x < 0) {x = -x;putchar('-');}
        if(x >= 10) out(x / 10);
        putchar('0' + x % 10);
    }
    int N;
    char strA[MAXN],strB[MAXN];
    int A[MAXN],B[MAXN],rev_A[MAXN],rev_B[MAXN],e[MAXN];
    
    int mul(int a,int b) {
        return 1LL * a * b % mo;
    }
    int inc(int a,int b) {
        return a + b >= mo ? a + b - mo : a + b;
    }
    int check1(int *a,int l,int r) {
        return mul(inc(a[r],mo - a[l - 1]),e[N - l]);
    }
    int check2(int *a,int l,int r) {
        return mul(inc(a[l],mo - a[r + 1]),e[r - 1]);
    }
    int Find(int *f,int *b,int s,int t) {
        if(s < 1 || t > N) return 0;
        int L = 0,R = min(N - t + 1,s);
        while(L < R) {
            int mid = (L + R + 1) >> 1;
            if(check1(f,s - mid + 1,s) == check2(b,t,t + mid - 1)) L = mid;
            else R = mid - 1;
        }
        return L;
    }
    void update(int &x,int y) {
        x = max(x,y);
    }
    void Solve() {
        read(N);
        scanf("%s%s",strA + 1,strB + 1);
        e[0] = 1;
        for(int i = 1 ; i <= N ; ++i) {
            e[i] = mul(e[i - 1],47);
        }
        for(int i = 1 ; i <= N ; ++i) {
            A[i] = mul(e[i - 1],strA[i] - 'A' + 1);
            A[i] = inc(A[i],A[i - 1]);
            B[i] = mul(e[i - 1],strB[i] - 'A' + 1);
            B[i] = inc(B[i],B[i - 1]);
        }
        for(int i = N ; i >= 1 ; --i) {
            rev_A[i] = mul(e[N - i],strA[i] - 'A' + 1);
            rev_A[i] = inc(rev_A[i],rev_A[i + 1]);
            rev_B[i] = mul(e[N - i],strB[i] - 'A' + 1);
            rev_B[i] = inc(rev_B[i],rev_B[i + 1]);
        }
        int ans = 1;
        for(int i = 2 ; i <= N - 1; ++i) {
            int L = Find(A,rev_A,i - 1,i + 1);
            int t = Find(A,rev_B,i - L - 1,i + L);
            update(ans,2 * L + 2 * t + 1);
        }
        for(int i = 1 ; i <= N - 1 ; ++i) {
            int L = Find(A,rev_A,i,i + 1);
            int t = Find(A,rev_B,i - L,i + L);
            update(ans,2 * L + 2 * t);
        }
        for(int i = 2 ; i <= N - 1; ++i) {
            int L = Find(B,rev_B,i - 1,i + 1);
            int t = Find(A,rev_B,i - L,i + L + 1);
            update(ans,2 * L + 2 * t + 1);
        }
        for(int i = 1 ; i <= N - 1; ++i) {
            int L = Find(B,rev_B,i,i + 1);
            int t = Find(A,rev_B,i - L + 1,i + L + 1);
            update(ans,2 * L + 2 * t);
        }
        out(ans);enter;
    }
    int main() {
    #ifdef ivorysi
        freopen("f1.in","r",stdin);
    #endif
        Solve();
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/ivorysi/p/9537146.html
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