两道相似KMP题

1.POJ 3450 Coporate Identity

这两题的解法都是枚举子串，然后匹配，像这种题目以后可以不用KMP来做，直接字符串自带的strstr函数搞定，如果字符串未出现，该函数返回NULL。

下面贴出其比较。

 

代码：（KMP版）（1360ms 888KB）

#include <iostream>

#include <cstdio>

#include <cstring>

#include <cmath>

#include <algorithm>

using namespace std;

#define N 4007



char ans[203],str[203];

char ss[N][203],tt[203];

int next[203];



void getnext(char *ss)

{

    int m = strlen(ss);

    next[0] = -1;

    int i = 0,j = -1;

    while(i<m)

    {

        if(j == -1 || ss[i] == ss[j])

            next[++i] = ++j;

        else

            j = next[j];

    }

}



int kmp(char *ss,char *tt)

{

    int n = strlen(ss);

    int m = strlen(tt);

    getnext(tt);

    int i = -1,j = -1;

    while(i<n && j<m)

    {

        if(j == -1 || ss[i] == tt[j])

            i++,j++;

        else

            j = next[j];

    }

    if(j == m)

        return 1;

    return 0;

}



int main()

{

    int n,i,len,j,k,lengh;

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

    {

        for(i=0;i<n;i++)

            scanf("%s",ss[i]);

        lengh = strlen(ss[0]);

        ans[0] = '\0';

        for(len=1;len<=lengh;len++)  //枚举长度

        {

            for(i=0;i<=lengh-len;i++)  //枚举起点

            {

                for(k=0,j=i;j<i+len;j++) //取出此字串

                    str[k++] = ss[0][j];

                str[k] = '\0';

                for(j=1;j<n;j++)

                {

                    if(!kmp(ss[j],str))

                        break;

                }

                if(j == n)

                {

                    if(strlen(ans) == len && strcmp(ans,str) > 0)

                        strcpy(ans,str);

                    if(strlen(ans) < len)

                        strcpy(ans,str);

                }

            }

        }

        if(ans[0] == '\0')

            cout<<"IDENTITY LOST\n";

        else

            cout<<ans<<endl;

    }

    return 0;

}

View Code

代码：（strstr函数版）（454ms 912KB）

#include <iostream>

#include <cstdio>

#include <cstring>

#include <cmath>

#include <algorithm>

#include <cstdlib>

using namespace std;

#define N 4007



char ans[203],str[203];

char ss[N][203],tt[203];

int main()

{

    int n,i,len,j,k,lengh;

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

    {

        for(i=0;i<n;i++)

            scanf("%s",ss[i]);

        lengh = strlen(ss[0]);

        ans[0] = '\0';

        for(len=1;len<=lengh;len++)  //枚举长度

        {

            for(i=0;i<=lengh-len;i++)  //枚举起点

            {

                for(k=0,j=i;j<i+len;j++) //取出此字串

                    str[k++] = ss[0][j];

                str[k] = '\0';

                for(j=1;j<n;j++)

                {

                    if(strstr(ss[j],str) == NULL)

                        break;

                }

                if(j == n)

                {

                    if(strlen(ans) == len && strcmp(ans,str) > 0)

                        strcpy(ans,str);

                    if(strlen(ans) < len)

                        strcpy(ans,str);

                }

            }

        }

        if(ans[0] == '\0')

            cout<<"IDENTITY LOST\n";

        else

            cout<<ans<<endl;

    }

    return 0;

}

View Code

 

2.POJ 3080 Blue Jeans

 

代码：（KMP版）（32ms 684KB）

#include <iostream>

#include <cstdio>

#include <cstring>

#include <cmath>

#include <algorithm>

using namespace std;

#define N 4007



char ans[65],str[65];

char ss[12][65],tt[65];

int next[65];



void getnext(char *ss)

{

    int m = strlen(ss);

    next[0] = -1;

    int i = 0,j = -1;

    while(i<m)

    {

        if(j == -1 || ss[i] == ss[j])

            next[++i] = ++j;

        else

            j = next[j];

    }

}



int kmp(char *ss,char *tt)

{

    int n = strlen(ss);

    int m = strlen(tt);

    getnext(tt);

    int i = -1,j = -1;

    while(i<n && j<m)

    {

        if(j == -1 || ss[i] == tt[j])

            i++,j++;

        else

            j = next[j];

    }

    if(j == m)

        return 1;

    return 0;

}



int main()

{

    int n,i,len,j,k,lengh,t;

    scanf("%d",&t);

    while(t--)

    {

        scanf("%d",&n);

        for(i=0;i<n;i++)

            scanf("%s",ss[i]);

        lengh = strlen(ss[0]);

        ans[0] = '\0';

        for(len=1;len<=lengh;len++)  //枚举长度

        {

            for(i=0;i<=lengh-len;i++)  //枚举起点

            {

                for(k=0,j=i;j<i+len;j++) //取出此字串

                    str[k++] = ss[0][j];

                str[k] = '\0';

                for(j=1;j<n;j++)

                {

                    if(!kmp(ss[j],str))

                        break;

                }

                if(j == n)

                {

                    if(strlen(ans) == len && strcmp(ans,str) > 0)

                        strcpy(ans,str);

                    if(strlen(ans) < len)

                        strcpy(ans,str);

                }

            }

        }

        if(ans[0] == '\0' || strlen(ans) < 3)

            cout<<"no significant commonalities\n";

        else

            cout<<ans<<endl;

    }

    return 0;

}

View Code

代码：（strstr函数版）（0ms 700KB）

#include <iostream>

#include <cstdio>

#include <cstring>

#include <cmath>

#include <algorithm>

using namespace std;

#define N 4007



char ans[65],str[65];

char ss[12][65];



int main()

{

    int n,i,len,j,k,lengh,t;

    scanf("%d",&t);

    while(t--)

    {

        scanf("%d",&n);

        for(i=0;i<n;i++)

            scanf("%s",ss[i]);

        lengh = strlen(ss[0]);

        ans[0] = '\0';

        for(len=1;len<=lengh;len++)  //枚举长度

        {

            for(i=0;i<=lengh-len;i++)  //枚举起点

            {

                for(k=0,j=i;j<i+len;j++) //取出此字串

                    str[k++] = ss[0][j];

                str[k] = '\0';

                for(j=1;j<n;j++)

                {

                    if(strstr(ss[j],str) == NULL)

                        break;

                }

                if(j == n)

                {

                    if(strlen(ans) == len && strcmp(ans,str) > 0)

                        strcpy(ans,str);

                    if(strlen(ans) < len)

                        strcpy(ans,str);

                }

            }

        }

        if(ans[0] == '\0' || strlen(ans) < 3)

            cout<<"no significant commonalities\n";

        else

            cout<<ans<<endl;

    }

    return 0;

}

View Code