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大意: 给定字符串$a$,$b$, $b$可以任选一段连续的区间删除, 要求最后$b$是$a$的子序列, 求最少删除区间长度.
删除一段连续区间, 那么剩余的一定是一段前缀和后缀. 判断是否是子序列可以用序列自动机, 最后双指针合并前缀与后缀的答案. #include <iostream> #include <sstream> #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' #define DB(a) ({REP(__i,1,m) cout<<a[__i]<<' ';hr;}) 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;} inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;} //head #ifdef ONLINE_JUDGE const int N = 1e6+10; #else const int N = 111; #endif char s[N], p[N]; int n, m, nxt[26][N], pre[26][N]; int L[N], R[N]; int main() { scanf("%s%s", s+1,p+1); n = strlen(s+1); m = strlen(p+1); REP(i,1,n) s[i]-='a'; REP(i,1,m) p[i]-='a'; REP(c,0,25) { nxt[c][n+1]=nxt[c][n+2]=n+1; PER(i,1,n) { if (s[i]==c) nxt[c][i]=i; else nxt[c][i]=nxt[c][i+1]; } REP(i,1,n) { if (s[i]==c) pre[c][i]=i; else pre[c][i]=pre[c][i-1]; } } L[1] = nxt[p[1]][1]; REP(i,2,m) L[i] = nxt[p[i]][L[i-1]+1]; R[m] = pre[p[m]][n]; PER(i,1,m-1) R[i] = pre[p[i]][R[i+1]-1]; int ans = 0, posR = 0, posL = 0; PER(i,1,m) if (R[i]) ans = m-i+1, posR = i; REP(i,1,m) p[i]+='a'; if (ans==m) return puts(p+1),0; R[m+1] = INF; int now = posR; REP(i,1,m) { while (R[now]<=L[i]) ++now; if (L[i]==n+1) break; if (ans<i+m-now+1) { ans = i+m-now+1; posL = i, posR = now; } } if (!ans) return puts("-"),0; REP(i,1,posL) putchar(p[i]); REP(i,posR,m) putchar(p[i]);hr; }
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