PKU 2192 Zipper

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PKU 2192 Zipper

http://acm.pku.edu.cn/JudgeOnline/problem?id=2192

晚上要讨论的题之一,先A一下

也是LCS的思想,关键是去找状态转移方程

AC CODE:

#include <iostream>

using namespace std;

#define maxlen 205

char a[maxlen]; //行

char b[maxlen]; //列

char str[maxlen*2];

bool recode[maxlen][maxlen] ;

// b a

int main()

{

int i,j,k,n,m,t,l;

while (scanf("%d",&t)!=EOF)

{

for (l=1;l<=t;l++)

{

memset(a+1,NULL,sizeof(a+1));

memset(b+1,NULL,sizeof(b+1));

memset(str+1,NULL,sizeof(str+1));

memset(recode,0,sizeof(recode));

scanf("%s",a+1);

scanf("%s",b+1);

scanf("%s",str+1);

int alen = strlen(a+1);

int blen = strlen(b+1);

recode[0][0] = 1;

for (i=0;i<=blen;i++) //列 b[]

{

for (j=0;j<=alen;j++) //行 a[]

{

if (i==0 && j==0)

{

continue;

}

if (recode[i-1][j] == 0 && recode[i][j-1] == 0 )

{

recode[i][j] = 0;

}

else if ((recode[i-1][j]&&str[i+j]==b[i]) || (recode[i][j-1]&&str[i+j]==a[j]))

{

recode[i][j] = 1;

}

//---------------------------------------------------

if (recode[blen][alen])

{

printf("Data set %d: yes/n",l);

}

else

{

printf("Data set %d: no/n",l);

}

return 0;

}