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 ;
}

Codeforces Round #425 (Div. 2) - B的更多相关文章

  1. Codeforces Round #425 (Div. 2)C

    题目连接:http://codeforces.com/contest/832/problem/C C. Strange Radiation time limit per test 3 seconds ...

  2. Codeforces Round #425 (Div. 2)

    A 题意:给你n根棍子,两个人每次拿m根你,你先拿,如果该谁拿的时候棍子数<m,这人就输,对手就赢,问你第一个拿的人能赢吗 代码: #include<stdio.h>#define ...

  3. Codeforces Round #425 (Div. 2) Problem D Misha, Grisha and Underground (Codeforces 832D) - 树链剖分 - 树状数组

    Misha and Grisha are funny boys, so they like to use new underground. The underground has n stations ...

  4. Codeforces Round #425 (Div. 2) Problem C Strange Radiation (Codeforces 832C) - 二分答案 - 数论

    n people are standing on a coordinate axis in points with positive integer coordinates strictly less ...

  5. Codeforces Round #425 (Div. 2) Problem B Petya and Exam (Codeforces 832B) - 暴力

    It's hard times now. Today Petya needs to score 100 points on Informatics exam. The tasks seem easy ...

  6. Codeforces Round #425 (Div. 2) Problem A Sasha and Sticks (Codeforces 832A)

    It's one more school day now. Sasha doesn't like classes and is always bored at them. So, each day h ...

  7. Codeforces Round #425 (Div. 2) B. Petya and Exam(字符串模拟 水)

    题目链接:http://codeforces.com/contest/832/problem/B B. Petya and Exam time limit per test 2 seconds mem ...

  8. Codeforces Round #425 (Div. 2))——A题&&B题&&D题

    A. Sasha and Sticks 题目链接:http://codeforces.com/contest/832/problem/A 题目意思:n个棍,双方每次取k个,取得多次数的人获胜,Sash ...

  9. Codeforces Round #425 (Div. 2) B - Petya and Exam

    地址:http://codeforces.com/contest/832/problem/B 题目: B. Petya and Exam time limit per test 2 seconds m ...

  10. Codeforces Round #425 (Div. 2) C - Strange Radiation

    地址:http://codeforces.com/contest/832/problem/C 题目: C. Strange Radiation time limit per test 3 second ...

随机推荐

  1. 基于Cesium的Web3d实现二三维室内外一体化融合

    https://cesiumjs.org/Cesium/Build/Apps/Sandcastle/index.html?src=3D%20Models.html 三维建模 BIM

  2. LUOGU P2569 [SCOI2010]股票交易(单调队列优化dp)

    传送门 解题思路 不难想一个\(O(n^3)\)的\(dp\),设\(f_{i,j}\)表示第\(i\)天,手上有\(j\)股的最大收益,因为这个\(dp\)具有单调性,所以\(f_i\)可以贪心的直 ...

  3. 3D Computer Grapihcs Using OpenGL - 03 OpenGL Buffer Data

    本节绘制一个三角形,并讲解Buffer Object-缓冲对象 OpenGL的窗口坐标 屏幕中心为坐标原点,横向朝右为x正方向,纵向朝上为y正方向,最大值最小值分别为1,-1. Buffer Obje ...

  4. 运行job检验单元测试覆盖率

    http://ns.jenkins.baidu.com/user/anyixing/my-views/view/Map_ut/job/poi-zhunru/ 1在http://ns.jenkins.b ...

  5. Linux下使用 ipset 封大量IP及ipset参数说明

    转载Linux下使用 ipset 封大量IP及ipset参数说明 Linux使用iptables封IP,是常用的应对网络攻击的方法,但要封禁成千上万个IP,如果添加成千上万条规则,对机器性能影响较大, ...

  6. DAY 4 下午

    一些图论的知识(主要补充一下之前不了解的和比较重要) 竞赛图:完全图上的边加方向 仙人掌:每一条边至多属于一个环 前序:中左右 中序:左中右 后序:左右中 先加进去无向边 把每一个联通块看成一个大点 ...

  7. fedora23帮定键盘系统操作快捷键

    在All settings -> keyboard 主要是以super为主, 然后有 super+ shift+...虽然感觉用 ctrl+super+... 来组合更方便, 但是用 shift ...

  8. cmd 编码修改。 牛阿。 解决问题

    http://jingyan.baidu.com/article/e75aca85440f01142edac636.html 命令窗口修改编码,CMD编码修改方法 听语音 | 浏览:9696 | 更新 ...

  9. github.com/oschwald/maxminddb-golang 安装报错

    安装 maxminddb-golang错误: dill@ubuntu-vm:~/workspace/go/src/github.com$ go get github.com/oschwald/maxm ...

  10. sqoop应用

    1.导入数据(将mysql(rdbms)的表的数据导入到hdfs) 1.1.全部导入(注意空格) sqoop import \ --connect jdbc:mysql://192.168.159.1 ...