【USACO】namenum

//开始傻×了 受题目形容的误导 一心想生成所有可能的 字符串组合 之后查找非常慢
//听了同学的 将5000个dict里的字符串 转换成char型数组（不能直接用int 会越界）直接用输入的数据对着转换后的数据查找就可以了 

//现在的程序还是太复杂了 没有必要提前把dict中的字符串都存起来 取一个判断一个就好了
#include <iostream>
#include <fstream>
#include <vector>
#include <algorithm>
#include <cstring>
using namespace std;

typedef struct
{
    char num[];
    char name[];
}MYDICT;
vector<MYDICT> dict; //vector<>中 最好不要用指针 因为vector对指针是不会复制的 如果之后用指针做了别的操作 vector里的值就都变了

int transName2Num(char * name, char * num)
{
    int len = strlen(name);
    int i;
    int n;
    for(i = , n = ; i < len; i++)
    {
        switch(name[i])
        {
        case 'A':
        case 'B':
        case 'C': num[i] = '';break;
        case 'D':
        case 'E':
        case 'F': num[i] = '';break;
        case 'G':
        case 'H':
        case 'I': num[i] = '';break;
        case 'J':
        case 'K':
        case 'L': num[i] = '';break;
        case 'M':
        case 'N':
        case 'O': num[i] = '';break;
        case 'P':
        case 'R':
        case 'S': num[i] = '';break;
        case 'T':
        case 'U':
        case 'V': num[i] = '';break;
        case 'W':
        case 'X':
        case 'Y': num[i] = '';break;
        default: break;
        }
    }
    num[len] = '\0';
    return ;
}

int main()
{
    MYDICT tmp;
    int flag = ;
    FILE * intxt, *out;
    char name[];
    char num[];
    char name3[];
    fstream in;
    in.open("namenum.in", ios::in);
    intxt = fopen("dict.txt", "r");
    out = fopen("namenum.out", "w");

    char innum[];
    in >> innum;
    do
    {
        fscanf(intxt,"%s", name);
        transName2Num(name, num);
        strcpy(tmp.name,name);
        strcpy(tmp.num,num);
        dict.push_back(tmp);
    }while(strcmp(name, "ZYTA") != );

    vector<MYDICT>::const_iterator it2;

    for(it2 = dict.begin(); it2 < dict.end(); it2++)
    {
        strcpy(name3,(*it2).num);
        if(strcmp(name3,innum) == )
        {
            fprintf(out, "%s\n", (*it2).name);
            flag = ;
        }
    }

    if(flag == )
    {
        fprintf(out, "NONE\n");
    }

    return ;
}

自己的代码还是太复杂了 下面是答案的 超精简

#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 = ;
    int numlen;
    char word[], num[], *p, *q, map[];
    int i, j;
    map['A'] = map['B'] = map['C'] = '';
    map['D'] = map['E'] = map['F'] = '';
    map['G'] = map['H'] = map['I'] = '';
    map['J'] = map['K'] = map['L'] = '';
    map['M'] = map['N'] = map['O'] = '';
    map['P'] = map['R'] = map['S'] = '';
    map['T'] = map['U'] = map['V'] = '';
    map['W'] = map['X'] = map['Y'] = '';
    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 == ) fprintf(out,"NONE\n");
    return ;
}