Two strings CodeForces - 762C (字符串,双指针)

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;
}

 

转载于:https://www.cnblogs.com/uid001/p/10826409.html