BNU OJ 第26303 题 Touchscreen Keyboard

　　BNU OJ第26303题Touchscreen Keyboard（题目链接）的解题报告。

　　原题如下：

Touchscreen Keyboard

Problem Description

Nowadays, people do not use hardware keyboards but touchscreens. Usually, they touch on the wrong letters with their chunky fingers, because screen space is precious and the letters therefore too small.

Usually, a spell checker runs after typing a word and suggests other words to select the correct spelling from. Your job is to order that list so that more likely words are on top. The typical touchscreen keyboard looks like this:

qwertyuiop
asdfghjkl
zxcvbnm

You should use the distance between the letters to type a word: the distance is the sum of the horizontal and vertical distance between the typed and proposed letter. Assume you typed a w, the distance to e is 1, while the distance to z is 3.

The typed word and the list of words from the spell checker all have the same length. The distance between two words is the sum of the letter distances. So the distance between ifpv and icpc is 3.

Input

The first line of the input specifies the number of test cases t (0 < t < 20). Each test case starts with a string and an integer l on one line. The string gives the word that was typed using the touchscreen keyboard, while l specifies the number of entries in the spell checker list (0 < l ≤ 10). Then follow l lines, each with one word of the spell checker list. You may safely assume that all words of one test case have the same length and no word is longer than 10 000 characters (only lowercase 'a' - 'z'). Furthermore, each word appears exactly once in the spell checker list on one test case.

Output

For each test case, print the list of words sorted by their distance ascending. If two words have the same distance, sort them alphabetically. Print the distance of each word in the same line.

Sample Input

2
ifpv 3
iopc
icpc
gcpc
edc 5
wsx
edc
rfv
plm
qed

Sample Output

icpc 3
gcpc 7
iopc 7
edc 0
rfv 3
wsx 3
qed 4
plm 17

 

 

 

 

　　先判断在键盘第几行，再判断在键盘的第几列。行差加上列差，即为所需误差，然后求和即可。

　　C++语言代码如下：

/***
*
*                     Touchscreen Keyboard
*
*
*                                                  叶剑飞
*                                                    2012年10月3日
*
*******************************************************************************/

#include 
#include 
#include 
#include 
#include 

using namespace std;

const char * first = "qwertyuiop";
const char * second = "asdfghjkll";
const char * third = "zxcvbnm";

typedef struct
{
    char word[10002];
    int sum;
} CORRECT;

int PosInFirst( const char ch )
{
    for ( int i = 0 ; first[i] != '\0' ; i ++ )
    {
        if ( ch == first[i] )
            return i + 1;
    }
    return 0;
}

int PosInSecond( const char ch )
{
    for ( int i = 0 ; second[i] != '\0' ; i ++ )
    {
        if ( ch == second[i] )
            return i + 1;
    }
    return 0;
}

int PosInThird( const char ch )
{
    for ( int i = 0 ; third[i] != '\0' ; i ++ )
    {
        if ( ch == third[i] )
            return i + 1;
    }
    return 0;
}

int FindError( const char a, const char b )
{
    int n;
    int pos_a, pos_b;
    if ( (pos_a = PosInFirst(a)) )
    {
        if ( (pos_b = PosInFirst(b)) )
        {
            n = pos_b - pos_a;
            if (n < 0 )
                n = -n;
        }
        else if ( (pos_b = PosInSecond(b)) )
        {
            n = pos_b - pos_a;
            if (n < 0 )
                n = -n;
            n ++;
        }
        else if ( (pos_b = PosInThird(b)) )
        {
            n = pos_b - pos_a;
            if (n < 0 )
                n = -n;
            n += 2;
        }
    }
    else if ( (pos_a = PosInSecond(a)) )
    {
        if ( (pos_b = PosInFirst(b)) )
        {
            n = pos_b - pos_a;
            if (n < 0 )
                n = -n;
            n ++;
        }
        else if ( (pos_b = PosInSecond(b)) )
        {
            n = pos_b - pos_a;
            if (n < 0 )
                n = -n;
        }
        else if ( (pos_b = PosInThird(b)) )
        {
            n = pos_b - pos_a;
            if (n < 0 )
                n = -n;
            n ++;
        }
    }
    else if ( (pos_a = PosInThird(a)) )
    {
        if ( (pos_b = PosInFirst(b)) )
        {
            n = pos_b - pos_a;
            if (n < 0 )
                n = -n;
            n += 2;
        }
        else if ( (pos_b = PosInSecond(b)) )
        {
            n = pos_b - pos_a;
            if (n < 0 )
                n = -n;
            n ++;
        }
        else if ( (pos_b = PosInThird(b)) )
        {
            n = pos_b - pos_a;
            if (n < 0 )
                n = -n;
        }
    }
    else
        assert( false );
    return n;
}

bool cmp( const CORRECT & a, const CORRECT & b )
{
    if ( a.sum == b.sum )
    {
        if ( strcmp( a.word, b.word ) < 0 )
            return true;
        else
            return false;
    }
    else
        return a.sum < b.sum;
}

int main (void)
{
    int i, j;
    int n, m;
    char wrong[10002];
    CORRECT correct[16];
    scanf( "%d", &n );
    while ( n -- )
    {
        scanf( "%s%d", wrong, &m );
        for ( i = 0 ; i < m  ; i ++ )
        {
            correct[i].sum = 0;
            scanf( "%s", correct[i].word );
            for ( j = 0; correct[i].word[j] != '\0' ; j ++ )
                correct[i].sum += FindError( wrong[j], correct[i].word[j] );
        }
        sort( correct, correct+m, cmp );
        for ( i = 0 ; i < m  ; i ++ )
            printf( "%s %d\n", correct[i].word, correct[i].sum );
    }
    return 0;
}

转载于:https://www.cnblogs.com/yejianfei/archive/2012/10/04/2711215.html