poj 1056 IMMEDIATE DECODABILITY(KMP)

题目链接：http://poj.org/problem?id=1056

思路分析：检测某字符串是否为另一字符串的前缀，数据很弱，可以使用暴力解法。这里为了练习KMP算法使用了KMP算法。

 

代码如下：

#include <iostream>
using namespace std;

const int N = 10;
const int Len = 20;
char A[N][Len];
int Next[N][Len];

void get_nextval( char P[], int Next[] )
{
    int i = 0, j = -1;
    int PLen = strlen(P);

    Next[0] = -1;
    while ( i < PLen - 1 )
    {
        if ( j == -1 || P[i] == P[j] )
        {
            i++;
            j++;
            if ( P[i] == P[j] )
                Next[i] = j;
            else
                Next[i] = Next[j];
        }
        else
            j = Next[j];
    }
}

int KMP_Matcher( char T[], char P[], int Next[] )
{
    int i = 0, j = 0;
    int TLen = strlen( T );
    int PLen = strlen( P );

    while ( i < TLen && j < PLen )
    {
        if ( j == -1 || T[i] == P[j] )
        {
            i++;
            j++;
        }
        else
            j = Next[j];
    }

    if ( j == PLen )
        return i - j;
    else
        return -1;
}


int main( )
{
    int Count = 0, flag = -1, n = 0;

    while ( scanf( "%s
", A[Count] ) != EOF )
    {
        if ( A[Count][0] == '9' )
        {
            n++;
            for( int i = 0; i < Count; ++i )
                get_nextval( A[i], Next[i] );

            for ( int i = 0; i < Count - 1; ++i )
                for ( int j = i + 1; j < Count; ++j )
                {
                    if ( flag == 0 )
                        break;
                    flag = KMP_Matcher( A[j], A[i], Next[i] );
                }

            if ( flag == 0 )
                printf( "Set %d is not immediately decodable
", n );
            else
                printf( "Set %d is immediately decodable
", n );
            
            Count = 0;
            flag = -1;
            continue;
        }

        Count++;
    }
    
    return 0;
}