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A newly opened detective agency is struggling with their limited intelligence to find out a secret
information passing technique among its detectives. Since they are new in this profession, they know
well that their messages will easily be trapped and hence modified by other groups. They want to guess
the intensions of other groups by checking the changed sections of messages. First they have to get the
length of longest match. You are going to help them.
Input
The input le may contain multiple test cases. Each case will contain two successive lines of string.
Blank lines and non-letter printable punctuation characters may appear. Each Line of string will be
no longer than 1000 characters. Length of each word will be less than 20 characters.
Output
For each case of input, you have to output a line starting with the case number right justified in a field
width of two, followed by the longest match as shown in the sample output. In case of at least one blank
line for each input output `
Blank!
Consider the non-letter punctuation characters as white-spaces.
SampleInput
This is a test.
test
Hello!
The document provides late-breaking information
late breaking.
SampleOutput
1. Length of longest match: 1
2. Blank!
3. Length of longest match: 2
这个题就是变形的最长公共子序列， 换汤不换药
最长公共子序列递推公式：
dp[i+1][j+1] = {
max(dp[i][j]+1, d[i][j+1], d[i+1][j]) (Si+1 = tj+1)
max(d[i][j+1], d[i+1][j]) 其他
}
#include <cstdio> #include <iostream> #include <cstring> #include <cstring> #include <cctype> using namespace std; const int maxn = 1003; string text[maxn], pattern[maxn]; int dp[maxn][maxn]; bool istrue(char c){ if((c >= ‘a‘ && c <= ‘z‘ )||(c >= ‘A‘ && c <= ‘Z‘) || (c >= ‘0‘ && c <= ‘9‘)){ return true; } return false; } void devide(string str[], string s, int &ans){ ans = 0; str[0].clear(); for(int i = 0; i < s.size(); i++){ if(istrue(s[i])){ str[ans] += s[i]; }else { if(str[ans].empty()){ continue; } ans++; str[ans].clear(); } } if(istrue(s[s.size()-1])) ans++; } int main() { string str1, str2; int num = 0, num1, num2;; while(!cin.eof()){ getline(cin, str1); getline(cin, str2); printf("%2d. ", ++num); if(str1.empty() || str2.empty()){ cout << "Blank!" << endl; }else { devide(text, str1, num1); devide(pattern, str2, num2); memset(dp, 0, sizeof(dp)); for(int i = 0; i < num1; i++){ for(int j = 0; j < num2; j++){ if(text[i] == pattern[j]){ dp[i+1][j+1] = dp[i][j] + 1; }else { dp[i+1][j+1] = max(dp[i+1][j], dp[i][j+1]); } } } cout << "Length of longest match: "<<dp[num1][num2] << endl; } } return 0; }
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原文地址：http://www.cnblogs.com/cshg/p/5723375.html
