牛客 109B 好位置 (字符串水题)

大意: 给定字符串$s1,s2$, 对于$s1$中所有与$s2$相等的子序列$t$, $t$在$s1$中的下标定义为好位置. 求$s1$是否所有位置都是好位置.

 

显然$s1$的前缀要与$s2$相等, 并且$s2$后缀连续相等的字符要与$s1$的后缀相等, 然后再判断下中间字符是否合法.

 

#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,n) cout<<a[__i]<<' ';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, P2 = 998244353, 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


const int N = 1e6+10;
int n, m, nxt[N][26], f[26];
char s1[N], s2[N];

int main() {
	scanf("%s%s", s1+1, s2+1);
	n = strlen(s1+1), m = strlen(s2+1);
	REP(i,1,m) if (s1[i]!=s2[i]) return puts("No"),0;
	for (int i=n,now=m; s2[now]==s2[m]; --now,--i) {
		if (s1[i]!=s2[m]) return puts("No"),0;
	}
	int x = s2[m];
	PER(i,1,m) {
		if (!f[s2[i]-'a']) f[s2[i]-'a'] = x;
		x = s2[i];
	}
	REP(i,0,25) nxt[n+1][i]=nxt[n+2][i]=n+1;
	PER(i,1,n) {
		memcpy(nxt[i],nxt[i+1],sizeof nxt[0]);
		nxt[i][s1[i]-'a']=i;
	}
	REP(i,m+1,n) if (s1[i]!=s2[m]) {
		if (!f[s1[i]-'a']) return puts("No"),0;
		if (nxt[i][f[s1[i]-'a']-'a']>n) return puts("No"),0;
	}
	puts("Yes");
}