Description

题库链接

给出一个长度为 \(n\) 的仅包含数字的字符串。 \(q\) 次询问，每次询问该串 \([a,b]\) 段内删去几个数能够使其不含 \(2016\) 的子串，但存在 \(2017\) 的子串。

\(4\leq n\leq 200000,1\leq q\leq 200000\)

Solution

考虑朴素的 \(DP\) 。我们记 \(f_{i,j}\) 为匹配到 \(i\) 这个位置拼成 "" "2" "20" "201" "2017" 的最小花费。显然这个是 \(O(nq)\) 的。

但对于每一位的转移是独立的。所以我们考虑用矩乘优化，线段树维护。

Code

//It is made by Awson on 2018.2.7

#include <bits/stdc++.h>

#define LL long long

#define dob complex<double>

#define Abs(a) ((a) < 0 ? (-(a)) : (a))

#define Max(a, b) ((a) > (b) ? (a) : (b))

#define Min(a, b) ((a) < (b) ? (a) : (b))

#define Swap(a, b) ((a) ^= (b), (b) ^= (a), (a) ^= (b))

#define writeln(x) (write(x), putchar('\n'))

#define lowbit(x) ((x)&(-(x)))

using namespace std;

const int N = 200000;

void read(int &x) {

    char ch; bool flag = 0;

    for (ch = getchar(); !isdigit(ch) && ((flag |= (ch == '-')) || 1); ch = getchar());

    for (x = 0; isdigit(ch); x = (x<<1)+(x<<3)+ch-48, ch = getchar());

    x *= 1-2*flag;

}

void print(int x) {if (x > 9) print(x/10); putchar(x%10+48); }

void write(int x) {if (x < 0) putchar('-'); print(Abs(x)); }

int n, q, a, b;

char ch[N+5];

struct mat {

    int a[5][5];

    mat() {memset(a, 127/3, sizeof(a)); }

    mat(int **_a) {for (int i = 0; i < 5; i++) for (int j = 0; j < 5; j++) a[i][j] = _a[i][j]; }

    mat operator * (const mat &b) const {

        mat ans;

        for (int i = 0; i < 5; i++)

            for (int j = 0; j < 5; j++)

                for (int k = 0; k < 5; k++)

                    ans.a[i][j] = Min(ans.a[i][j], a[i][k]+b.a[k][j]);

        return ans;

    }

};

struct Segment_tree {

    mat sgm[(N<<2)+5];

    #define lr(x) (x<<1)

    #define rr(x) (x<<1|1)

    void build(int o, int l, int r) {

        if (l == r) {

            for (int i = 0; i < 5; i++) sgm[o].a[i][i] = 0;

            if (ch[l] == '2') sgm[o].a[0][0] = 1, sgm[o].a[0][1] = 0;

            else if (ch[l] == '0') sgm[o].a[1][1] = 1, sgm[o].a[1][2] = 0;

            else if (ch[l] == '1') sgm[o].a[2][2] = 1, sgm[o].a[2][3] = 0;

            else if (ch[l] == '7') sgm[o].a[3][3] = 1, sgm[o].a[3][4] = 0;

            else if (ch[l] == '6') sgm[o].a[3][3] = 1, sgm[o].a[4][4] = 1;

            return;

        }

        int mid = (l+r)>>1;

        build(lr(o), l, mid), build(rr(o), mid+1, r);

        sgm[o] = sgm[lr(o)]*sgm[rr(o)];

    }

    mat query(int o, int l, int r, int a, int b) {

        if (a <= l && r <= b) return sgm[o];

        int mid = (l+r)>>1;

        if (b <= mid) return query(lr(o), l, mid, a, b);

        if (a > mid) return query(rr(o), mid+1, r, a, b);

        return query(lr(o), l, mid, a, b)*query(rr(o), mid+1, r, a, b);

    }

}T;

void work() {

    read(n), read(q); scanf("%s", ch+1);

    T.build(1, 1, n);

    while (q--) {

        read(a), read(b);

        mat tmp = T.query(1, 1, n, a, b);

        writeln(tmp.a[0][4] > n ? -1 : tmp.a[0][4]);

    }

}

int main() {

    work(); return 0;

}