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poj 1256 Anagram

题目链接:http://poj.org/problem?id=1256

 

思路:

    该题为含有重复元素的全排列问题;由于题目中字符长度较小,采用暴力法解决。

代码如下:

 

#include <iostream>#include <algorithm>using namespace std;const int MAX_N = 20;char P[MAX_N], A[MAX_N];char * SortAlp( char P[], int n ){    int Low[MAX_N], Upper[MAX_N];    int LowLen, UpperLen;    LowLen = UpperLen = 0;    for ( int i = 0; i < n; ++i )    {        if ( A <= P[i] && P[i] <= Z )            Upper[UpperLen++] = P[i];        else            Low[LowLen++] = P[i];    }    sort( Low, Low + LowLen );    sort( Upper, Upper + UpperLen );    int Index_L, Index_U;    Index_L = Index_U = 0;    for ( int j = 0; j < n; ++j )    {        if (Upper[Index_U] - A + a <= Low[Index_L]            && Index_U < UpperLen)            P[j] = Upper[Index_U++];        else            P[j] = Low[Index_L++];    }    return P;}void PrintPermutation( int n, char P[], char A[], int cur ){    int i, j;    if ( cur == n )    {        for ( i = 0; i < n; ++i )            printf( "%c", A[i] );        printf("\n");    }    else    {        for ( i = 0; i < n; ++i )        {            if ( !i || P[i] != P[i-1] )            {                int c1 = 0, c2 = 0;                for ( j = 0; j < cur; ++j )                    if ( A[j] == P[i] ) c1++;                for ( j = 0; j < n; ++j )                    if ( P[i] == P[j] ) c2++;                if ( c1 < c2 )                {                    A[cur] = P[i];                    PrintPermutation( n, P, A, cur + 1 );                }            }        }    }}int main(){    int n;    char P[MAX_N];    cin >> n;    for ( int i = 0; i < n; ++i )    {        cin >> P;        SortAlp( P, strlen(P) );        PrintPermutation( strlen(P), P, A, 0 );    }    return 0;}

 

poj 1256 Anagram