题目链接:https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=13&page=show_problem&problem=1041

LCS类型的题,不过并不是找common character,而是common word.就先把string处理成a list of word,然后再用LCS算法求common word。

代码如下:

 #include <iostream>
#include <math.h>
#include <stdio.h>
#include <cstdio>
#include <algorithm>
#include <string.h>
#include <cstring>
#include <queue>
#include <vector>
#include <functional>
#include <cmath>
#define SCF(a) scanf("%d", &a)
#define IN(a) cin>>a
#define FOR(i, a, b) for(int i=a;i<b;i++)
typedef long long Int;
using namespace std; int main()
{
char str1[], str2[];
vector<string> v1, v2;
int testCase = ;
int len1 = , len2 = ;
while (cin.getline(str1, ))
{
cin.getline(str2, );
int cnum = ;
char word[];
string wd;
len1 = ;
len2 = ;
for (int i = ; str1[i] != '\0'; i++)
{
len1++;
if ((str1[i] >= 'a' && str1[i] <= 'z') || (str1[i] >= 'A' && str1[i] >= 'Z') || (str1[i] >= '' && str1[i] <= ''))
{
word[cnum++] = str1[i];
}
else
{
if (cnum > )
{
word[cnum++] = '\0';
wd = string(word);
v1.push_back(wd);
}
cnum = ;
}
}
if (cnum > )
{
word[cnum++] = '\0';
wd = string(word);
v1.push_back(wd);
}
cnum = ;
for (int i = ; str2[i] != '\0'; i++)
{
len2++;
if ((str2[i] >= 'a' && str2[i] <= 'z') || (str2[i] >= 'A' && str2[i] >= 'Z') || (str2[i] >= '' && str2[i] <= ''))
{
word[cnum++] = str2[i];
}
else
{
if (cnum > )
{
word[cnum++] = '\0';
wd = string(word);
v2.push_back(wd);
}
cnum = ;
}
}
if (cnum > )
{
word[cnum++] = '\0';
wd = string(word);
v2.push_back(wd);
} int **match = new int*[v1.size() + ];
FOR(i, , v1.size() + )
match[i] = new int[v2.size() + ]; FOR(i, , v1.size() + )
match[i][] = ;
FOR(i, , v2.size() + )
match[][i] = ; FOR(i, , v1.size() + )
{
FOR(j, , v2.size() + )
{
if (v1[i - ] == v2[j - ])
match[i][j] = match[i - ][j - ] + ;
else
match[i][j] = max(match[i - ][j], match[i][j - ]);
}
}
if(len1== || len2==)
printf("%2d. Blank!\n", testCase++);
else
printf("%2d. Length of longest match: %d\n", testCase++, match[v1.size()][v2.size()]); while (!v1.empty())
v1.pop_back();
while (!v2.empty())
v2.pop_back(); FOR(i, , v1.size() + )
delete[] match[i];
delete[] match; }
return ;
}

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