New Year and Old Subsequence CodeForces - 750E (dp矩阵优化)

大意: 给定字符串, 每次询问区间[l,r]有子序列2017, 无子序列2016所需要删除的最小字符数

转移用矩阵优化一下, 要注意$(\mathbb{Z},min,+)$的幺元主对角线全0, 其余全无穷, 零元为全无穷

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '\n'
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
//head

const int N = 2e5+10;
int n, q;
char s[N];
struct _ {
	int v[5][5];
	_ operator * (const _ & rhs) const {
		_ r;
		memset(r.v,0x3f,sizeof r.v);
		REP(k,0,4) REP(i,0,4) REP(j,0,4) {
			r.v[i][j] = min(r.v[i][j],v[i][k]+rhs.v[k][j]);
		}
		return r;
	}
} mx[N<<2];
_ build(int o, int l, int r) {
	if (l==r) {
		memset(mx[o].v,0x3f,sizeof mx[o].v);
		REP(i,0,4) mx[o].v[i][i]=0;
		if (s[l]=='2') mx[o].v[0][0]=1,mx[o].v[0][1]=0;
		if (s[l]=='0') mx[o].v[1][1]=1,mx[o].v[1][2]=0;
		if (s[l]=='1') mx[o].v[2][2]=1,mx[o].v[2][3]=0;
		if (s[l]=='7') mx[o].v[3][3]=1,mx[o].v[3][4]=0;
		if (s[l]=='6') mx[o].v[3][3]=1,mx[o].v[4][4]=1;
		return mx[o];
	}
	return mx[o]=build(ls)*build(rs);
}
_ qry(int o, int l, int r, int ql, int qr) {
	if (ql<=l&&r<=qr) return mx[o];
	if (mid>=qr) return qry(ls,ql,qr);
	if (mid<ql) return qry(rs,ql,qr);
	return qry(ls,ql,qr)*qry(rs,ql,qr);
}

int main() {
	scanf("%d%d%s", &n, &q, s+1);
	build(1,1,n);
	REP(i,1,q) {
		int l, r;
		scanf("%d%d", &l, &r);
		int ans = qry(1,1,n,l,r).v[0][4];
		printf("%d\n", ans==INF?-1:ans);
	}
}