Subsequence（暴力）

最新推荐文章于 2021-11-25 10:10:00 发布

原创最新推荐文章于 2021-11-25 10:10:00 发布 · 218 阅读

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#暴力

本文介绍了一道ICPC中国南昌邀请赛预选赛的编程题，题目要求判断一系列字符串是否为给定字符串的子序列。提供了两种解题思路，一种是直接使用strchr函数进行暴力匹配，另一种是通过预处理字符位置来优化查询速度。

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The Preliminary Contest for ICPC China Nanchang National Invitational
题目链接：https://nanti.jisuanke.com/t/38232

Give a string S and N string T_i, determine whether T_i is a subsequence of S.
If ti is subsequence of S, print YES,else print NO.
If there is an array {K₁,K₂,K₃,⋯ ,K_m} so that 1≤K₁<K₂<K₃<⋯<K_m≤N and S_{k_i}=T_i(1≤i≤m), then T_i is a subsequence of S.

Input

The first line is one string S,length(S) ≤100000
The second line is one positive integer N,N≤100000
Then next nnn lines,every line is a string T_i, length(T_i) ≤1000

Output

Print N lines. If the i-th T_i is subsequence of S, print YES, else print NO.

样例输入

abcdefg
3
abc
adg
cba

样例输出

YES
YES
NO

直接暴力居然可以过？！ヽ(≧□≦)ノ难受了…果然要少用cin、cout？

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<string.h>
using namespace std;
char s[100005],t[1005];
int main()
{
	int n, flag, lens, lent;
	char *pos;
	scanf("%s", s);
	lens = strlen(s);
	scanf("%d", &n);
	while (n--)
	{
		scanf("%s", t);
		lent = strlen(t);
		pos = s;
		flag = 1;
		for (int i = 0; i < lent; i++)
		{
			pos = strchr(pos, t[i]);
			if (pos == NULL)
			{
				flag = 0;
				break;
			}
			pos++;
		}
		if (flag)
			printf("YES\n");
		else printf("NO\n");
	}
	return 0;
}

优化的做法（因为之前用cin、cout疯狂T）

#include<iostream>
#include<algorithm>
#include<string>
#include<cstring>
using namespace std;
int pos[100005][30];
char s[100005], t[1005];
int main()
{
	scanf("%s",s);
	int len = strlen(s),n,temp,flag;
	for (int i = len; i > 0; i--)
	{
		for (int j = 0; j < 26; j++)
		{
			pos[i - 1][j] = pos[i][j];
		}
		pos[i-1][s[i] - 'a'] = i+1;
	}
	pos[0][s[0] - 'a'] = 1;
	scanf("%d", &n); 
	while (n--)
	{
		scanf("%s", t);
		len = strlen(t);
		flag = 1;
		temp = pos[0][t[0]-'a'];
		for (int i = 1; i < len; i++)
		{
			if (temp <= 0)
			{
				flag = 0;
				break;
			}
			temp = pos[temp-1][t[i]-'a'];
		}
		if (flag && temp != 0)
			printf("YES\n");
		else
			printf("NO\n");
	}
	return 0;
}