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); } }