hdu 6170 正则表达式 dp
传送门
Two strings
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 463 Accepted Submission(s): 171
Problem Description
Giving two strings and you should judge if they are matched.
The first string contains lowercase letters and uppercase letters.
The second string contains lowercase letters, uppercase letters, and special symbols: “.” and “*”.
. can match any letter, and * means the front character can appear any times. For example, “a.b” can match “acb” or “abb”, “a*” can match “a”, “aa” and even empty string. ( “*” will not appear in the front of the string, and there will not be two consecutive “*”.
The first string contains lowercase letters and uppercase letters.
The second string contains lowercase letters, uppercase letters, and special symbols: “.” and “*”.
. can match any letter, and * means the front character can appear any times. For example, “a.b” can match “acb” or “abb”, “a*” can match “a”, “aa” and even empty string. ( “*” will not appear in the front of the string, and there will not be two consecutive “*”.
Input
The first line contains an integer T implying the number of test cases. (T≤15)
For each test case, there are two lines implying the two strings (The length of the two strings is less than 2500).
For each test case, there are two lines implying the two strings (The length of the two strings is less than 2500).
Output
For each test case, print “yes” if the two strings are matched, otherwise print “no”.
Sample Input
3 aa a* abb a.* abb aab
Sample Output
yes yes no
Source
2017 Multi-University Training Contest - Team 9
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题意
给出原串与匹配串，问能否匹配原串中所有的字符。
思路
如果这是一个标准的正则匹配，是不是可以直接用语言特性了呢？
我们设原串为 a ，匹配串为 b ， dp[i][j] 代表 b[1..i] 与 a[1..j] 是否匹配成功。
显然 dp[0][0] = true ，
对于其他情况：
-
如果
b[i] == '.'，则此时a[j]可以是任意字符，dp[i][j]由dp[i-1][j-1]转移而来。 -
如果
a[j] == b[i]，同样dp[i][j]由dp[i-1][j-1]转移而来。 -
如果
b[i] == '*'，假设该*最终可以匹配 0 位，则dp[i][j]状态从dp[i-2][j]转移而来，假设最终匹配 1 位，则从dp[i-1][j]转移而来；假如
a[1..j-1]与b[1..i-1]已成功匹配，并且a[j-1] == a[j]，显然当前的*可以继续匹配这一个字符，因此dp[i][j] = true；假如
a[1..j-1]与b[1..i]已成功匹配（当前*已成功匹配若干位），且a[j-1] == a[j]，则可以继续匹配这一个字符，因此dp[i][j] = true。 -
特别的，如果
b[2] == '*'，则dp[ALL][0] = true。
//china no.1
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include
#include
#include
#include 另：regex大佬
#include
#include
#include
#include
#include
#include
using namespace std;
const int MAXN=2600;
string s,p;
int main()
{
int T;
cin>>T;
while(T--)
{
cin>>s>>p;
string s1=".*";
string s2="(a*|b*|c*|d*|e*|f*|g*|h*|i*|j*|k*|l*|m*|n*|o*|p*|q*|r*|s*|t*|u*|v*|w*|x*|y*|z*"
"|A*|B*|C*|D*|E*|F*|G*|H*|I*|J*|K*|L*|M*|N*|O*|P*|Q*|R*|S*|T*|U*|V*|W*|X*|Y*|Z*)";
auto pos=p.find(s1);
while(pos!=string::npos)
{
p.replace(pos,2,s2);
pos=p.find(s1,pos+157);
}
regex pat(p);
if(regex_match(s,pat)) cout<<"yes"<