Codeforces Round #425 (Div. 2) - B
题目链接:http://codeforces.com/contest/832/problem/B
题意:给定一个好字母集合(只有小写字母,除了这些外其余都是坏字母集合),给定一个匹配模式串,
模式串只包含小写字母,'*','?'; 其中'?'可以替换成“好字母集合”中的任意一个字母
‘*’可以替换成空字符串,坏字母组成的字符串
然后给你n个字符串,问你这n个字符串哪些满足给定的匹配模式。
思路:按照题意模拟,或者直接写个正则表达式来匹配即可。
#define _CRT_SECURE_NO_DEPRECATE
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<string>
#include<queue>
#include<vector>
#include<time.h>
#include<cmath>
using namespace std;
typedef long long int LL;
const int MAXN = 1e5 + ;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + ;
char gchar[], pattern[MAXN], str[MAXN];
int vis[];
int main(){
#ifdef kirito
freopen("in.txt", "r", stdin);
freopen("out.txt", "w", stdout);
#endif
int n, slen, plen, glen;
while (~scanf("%s", gchar)){
scanf("%s", pattern); scanf("%d", &n);
plen = strlen(pattern); glen = strlen(gchar);
bool flag = false;
for (int i = ; i < plen; i++){
if (pattern[i] == '*'){
flag = true; break;
}
}
memset(vis, , sizeof(vis));
for (int i = ; i < glen; i++){
vis[gchar[i] - 'a'] = ;
}
for (int i = , j, k; i <= n; i++){
scanf("%s", str);
slen = strlen(str);
bool res = true;
if (plen - == slen&&flag){
int pos = ;
for (j = ; j < slen&&res; j++, pos++){
if (pattern[pos] == '*'){ //换成空字符串
j--;
}
else if (pattern[pos] == '?'){
if (!vis[str[j] - 'a']){
res = false;
}
}
else{
if (pattern[pos] != str[j]){
res = false;
}
}
}
}
else if (plen <= slen){
int pos = , pslen = slen - plen;
if (pslen&&!flag){
res = false;
}
for (j = ; j < slen&&res; j++, pos++){
if (pattern[pos] == '*'){ //换成坏字母组成的字符串
for (k = j; k <= j + pslen; k++){
if (vis[str[k] - 'a']){
res = false; break;
}
}
j = j + pslen;
}
else if (pattern[pos] == '?'){
if (!vis[str[j] - 'a']){
res = false;
}
}
else{
if (pattern[pos] != str[j]){
res = false;
}
}
}
}
else{
res = false;
}
printf(res ? "YES\n" : "NO\n");
}
}
return ;
}
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