[Codeforces 750E]New Year and Old Subsequence

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。

(4leq nleq 200000,1leq qleq 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('
'))
#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;
}