USACO section 1.2.3 Name That Number

1. 比较简单，但是用到了文件读写，终于明白了给的test代码中的fout和fin是什么意思了，哈哈；

2. 如果按照一般的思路，肯定会超时，所以把dict转换成数字，这样的算法效率是固定的

以下是代码：

/*
ID: dollar4
PROG: namenum
LANG: C++
*/
#include <iostream>
#include <fstream>
#include <string>
#include <algorithm>
#include <cstring>

using namespace std;
char get_num(char a)
{
    switch (a)
    {
    case 'A' :
        return '2';
    case 'B' :
        return '2';
    case 'C' :
        return '2';

    case 'D' :
        return '3';
    case 'E' :
        return '3';
    case 'F' :
        return '3';

    case 'G' :
        return '4';
    case 'H' :
        return '4';
    case 'I' :
        return '4';

    case 'J' :
        return '5';
    case 'K' :
        return '5';
    case 'L' :
        return '5';

    case 'M' :
        return '6';
    case 'N' :
        return '6';
    case 'O' :
        return '6';

    case 'P' :
        return '7';
    case 'R' :
        return '7';
    case 'S' :
        return '7';

    case 'T' :
        return '8';
    case 'U' :
        return '8';
    case 'V' :
        return '8';

    case 'W' :
        return '9';
    case 'X' :
        return '9';
    case 'Y' :
        return '9';
    }
}
int main()
{
    ofstream fout ("namenum.out");
    ifstream fin ("namenum.in");
    string str, fstr, cstr;
    int i, cnt = 0;
    ifstream dictfile("dict.txt");
    fin >> str;
    while (dictfile >> fstr)
    {
        cstr = "";
        for (unsigned int k = 0; k < fstr.length(); k++)
        {
            cstr += get_num(fstr[k]);
        }
//        cout << cstr << endl;
        if (str == cstr)
        {
            fout << fstr << endl;
            cnt++;
        }
    }
    if (cnt == 0)
        fout << "NONE" << endl;
    return 0;
}

以下是官方参考代码：

/*Here is Argentina competitor's Michel Mizrah's solution using the first method with a binary search. While it is blazingly fast, it does have the disadvantage of some fairly tricky coding in the binary search routine. A single off-by-one error would doom a program in a contest.
*/
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char num[12],sol[12];
char dict[5000][13];
int nsolutions = 0;
int nwords;
int maxlen;
FILE *out;

void calc (int charloc, int low, int high) {
    if (charloc == maxlen) {
        sol[charloc] = '\0';
        for (int x = low; x < high; x++) {
            if (strcmp (sol, dict[x]) == 0) {
                fprintf (out, "%s\n", sol);
                nsolutions++;
            }
        }
        return;
   }
   if (charloc > 0) {
        for (int j=low; j <= high; j++){
            if (sol[charloc-1] == dict[j][charloc-1]) {
                low=j;
                while (sol[charloc-1] == dict[j][charloc-1])
                    j++;
                high=j;
                break;
            }
            if (j == high) return;
        }
    }
    if (low > high) return;
    switch(num[charloc]){
      case '2':sol[charloc] = 'A'; calc(charloc+1,low,high);
               sol[charloc] = 'B'; calc(charloc+1,low,high);
               sol[charloc] = 'C'; calc(charloc+1,low,high);
               break; 
      case '3':sol[charloc] = 'D'; calc(charloc+1,low,high);
               sol[charloc] = 'E'; calc(charloc+1,low,high);
               sol[charloc] = 'F'; calc(charloc+1,low,high);
               break; 
      case '4':sol[charloc] = 'G'; calc(charloc+1,low,high);
               sol[charloc] = 'H'; calc(charloc+1,low,high);
               sol[charloc] = 'I'; calc(charloc+1,low,high);
               break; 
      case '5':sol[charloc] = 'J'; calc(charloc+1,low,high);
               sol[charloc] = 'K'; calc(charloc+1,low,high);
               sol[charloc] = 'L'; calc(charloc+1,low,high);
               break; 
      case '6':sol[charloc] = 'M'; calc(charloc+1,low,high);
               sol[charloc] = 'N'; calc(charloc+1,low,high);
               sol[charloc] = 'O'; calc(charloc+1,low,high);
               break; 
      case '7':sol[charloc] = 'P'; calc(charloc+1,low,high);
               sol[charloc] = 'R'; calc(charloc+1,low,high);
               sol[charloc] = 'S'; calc(charloc+1,low,high);
               break; 
      case '8':sol[charloc] = 'T'; calc(charloc+1,low,high);
               sol[charloc] = 'U'; calc(charloc+1,low,high);
               sol[charloc] = 'V'; calc(charloc+1,low,high);
               break; 
      case '9':sol[charloc] = 'W'; calc(charloc+1,low,high);
               sol[charloc] = 'X'; calc(charloc+1,low,high);
               sol[charloc] = 'Y'; calc(charloc+1,low,high);
               break;
   }
}

int main(){
    FILE *in=fopen ("namenum.in", "r");
    FILE *in2=fopen ("dict.txt", "r");
    int j;
    out=fopen ("namenum.out","w");
    for (nwords = 0; fscanf (in2, "%s", &dict[nwords++]) != EOF; )
        ;
    fscanf (in, "%s",&num);
    maxlen = strlen(num);
    calc (0, 0, nwords);
    if (nsolutions == 0) fprintf(out,"NONE\n");
    return 0;
}

另一个官方代码：

/*The solution below might be considered to be a bit more straightforward: no tricky offsets, no +1 or -1, no knowledge about character values. The lines of actual code in this solution are minimal.*/

This is the sort of program that might work reliably the first time and every time. The only tricky part is knowing that scanf will yield string without a newline on the end:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int main() {
    FILE *in = fopen ("namenum.in", "r");
    FILE *in2 = fopen ("dict.txt", "r");
    FILE *out = fopen ("namenum.out","w");
    int nsolutions = 0;
    int numlen;
    char word[80], num[80], *p, *q, map[256];
    int i, j;
    map['A'] = map['B'] = map['C'] = '2';
    map['D'] = map['E'] = map['F'] = '3';
    map['G'] = map['H'] = map['I'] = '4';
    map['J'] = map['K'] = map['L'] = '5';
    map['M'] = map['N'] = map['O'] = '6';
    map['P'] = map['R'] = map['S'] = '7';
    map['T'] = map['U'] = map['V'] = '8';
    map['W'] = map['X'] = map['Y'] = '9';
    fscanf (in, "%s",num);
    numlen = strlen(num);
    while (fscanf (in2, "%s", word) != EOF) {
        for (p=word, q=num; *p && *q; p++, q++) {
            if (map[*p] != *q)
                break;
        }
        if (*p == '\0' && *q == '\0') {
            fprintf (out, "%s\n", word);
            nsolutions++;
        }
    }
    if (nsolutions == 0) fprintf(out,"NONE\n");
    return 0;
}