POJ_2250_LCS

//============================================================================
// Name : POJ_2250_LCS.cpp
// Author : tiger
// Version :
// Copyright : 最长公共子串
// Description : 将每个单词看做一个字符，求最大公共子串，关键是记录路径
// 用record[][]记录，当s1[i]==s2[j]时，cmp[i][j]得值来自于cmp[i-1][j-1]+1
// ,则record[i][j] = 2;当s1[i]!=s2[j]时,如果cmp[i][j]得值来自于
// cmp[i-1][j],则record[i][j] = 3;如果cmp[i][j]得值来自于
// cmp[i][j-1],则record[i][j] = 1;如果cmp[i][j]得值来自于
// cmp[i-1][j-1],则record[i][j] = 2;
// 这样从后向前推，就能找到所求，数据较弱
//============================================================================

#include <iostream>
#include <string>
#include <stdio.h>
using namespace std;
char s1[101][31];
char s2[101][31];
int cmp[101][101];
int record[101][101];
int len1,len2;
int LCS(int *maxi,int *maxj)
{
memset(cmp,0,sizeof(cmp));

int i,j,k;
int max = 0;
for(i = 0; i < len1; i++)
{
for(j = 0; j < len2; j++)
{
if(strcmp(s1[i],s2[j]) == 0 )
{
if(i ==0 || j == 0)
{
cmp[i][j] = 1;
}
else
{

cmp[i][j] = cmp[i-1][j-1] +1;
}
record[i][j] = 2;
if(cmp[i][j] > max)
{
max = cmp[i][j];
*maxi = i;
*maxj = j;
}

}else
{
k = 0;
if(i > 0 && cmp[i-1][j] > k)
{

k = cmp[i-1][j];
record[i][j] = 3;
}
if(j > 0 && cmp[i][j-1] > k)
{
k = cmp[i][j-1];
record[i][j] = 1;

}
if(j > 0 && i > 0 && cmp[i-1][j-1] > k)
{

k = cmp[i-1][j-1];
record[i][j] = 2;
}
cmp[i][j] = k;

}
// printf("%d ",record[i][j]);
}
//printf("/n");
}
return max;//返回LCS长度
}

int main() {

freopen("in","r",stdin);
int i,j,k,max;
int ans[101];
while(scanf("%s",s1[0]) != EOF)
{
len1 = 1;
while(scanf("%s",s1[len1]) && s1[len1][0] != '#')
{
len1++;
}
len2 = 0;
while(scanf("%s",s2[len2]) && s2[len2][0] != '#')
{
len2++;
}

max = LCS(&i,&j);
//从后向前查找LCS，并记录其在s2中的下标于ans[]
k = 0;
for( ;i >= 0 && k < max; )
{
for( ; j >= 0 && k < max; )
{
if(record[i][j] == 2)
{
ans[k++] = j;

i--;j--;
}
else
if(record[i][j] == 3)
{
i--;
}else
j--;
}
}
while(k > 0)
{
k--;
printf("%s ",s2[ans[k]]);
}
printf("/n");

}
return 0;
}