POJ1002 487-3279

题目：

Description

Businesses like to have memorable telephone numbers. One way to make a telephone number memorable is to have it spell a memorable word or phrase. For example, you can call the University of Waterloo by dialing the memorable TUT-GLOP. Sometimes only part of the number is used to spell a word. When you get back to your hotel tonight you can order a pizza from Gino's by dialing 310-GINO. Another way to make a telephone number memorable is to group the digits in a memorable way. You could order your pizza from Pizza Hut by calling their ``three tens'' number 3-10-10-10. 
The standard form of a telephone number is seven decimal digits with a hyphen between the third and fourth digits (e.g. 888-1200). The keypad of a phone supplies the mapping of letters to numbers, as follows: 
A, B, and C map to 2 
D, E, and F map to 3 
G, H, and I map to 4 
J, K, and L map to 5 
M, N, and O map to 6 
P, R, and S map to 7 
T, U, and V map to 8 
W, X, and Y map to 9 
There is no mapping for Q or Z. Hyphens are not dialed, and can be added and removed as necessary. The standard form of TUT-GLOP is 888-4567, the standard form of 310-GINO is 310-4466, and the standard form of 3-10-10-10 is 310-1010. 
Two telephone numbers are equivalent if they have the same standard form. (They dial the same number.) 
Your company is compiling a directory of telephone numbers from local businesses. As part of the quality control process you want to check that no two (or more) businesses in the directory have the same telephone number. 

Input

The input will consist of one case. The first line of the input specifies the number of telephone numbers in the directory (up to 100,000) as a positive integer alone on the line. The remaining lines list the telephone numbers in the directory, with each number alone on a line. Each telephone number consists of a string composed of decimal digits, uppercase letters (excluding Q and Z) and hyphens. Exactly seven of the characters in the string will be digits or letters. 

Output

Generate a line of output for each telephone number that appears more than once in any form. The line should give the telephone number in standard form, followed by a space, followed by the number of times the telephone number appears in the directory. Arrange the output lines by telephone number in ascending lexicographical order. If there are no duplicates in the input print the line: 
No duplicates. 

Sample Input

12
4873279
ITS-EASY
888-4567
3-10-10-10
888-GLOP
TUT-GLOP
967-11-11
310-GINO
F101010
888-1200
-4-8-7-3-2-7-9-
487-3279

Sample Output

310-1010 2
487-3279 4
888-4567 3

思路：

题目很简单，将输入的电话号码转成字母，并且找出重复出现的电话号码和重复次数。主要就在于大数据量的处理问题。注意题目的时间限制为2000ms，我下面的代码AC的时间是1375ms，真是险过啊。。。看来自己的姿势水平还是不行啊

代码：(memory:5200k, time 1375ms)

#include<iostream>
#include<map>
#include<string>

using namespace std;

int main(){
	int num;
	map<string,int> phones;
	int mapping[26]={2,2,2,3,3,3,4,4,4,5,5,5,6,6,6,7,-1,7,7,8,8,8,9,9,9,-1};
	map<string, int>::iterator iter;
	cin>>num;
	for(int i=0;i<num;i++){
		string phone="";
		string a;
		cin>>a;
		for(unsigned int i=0;i<a.length();i++){
			if(a[i]>='A'&&a[i]<='Z'){
                phone+=mapping[a[i]-'A']+'0';
			}
			else if(a[i]>='0'&&a[i]<='9'){
				phone+=a[i];
	        }
		}
		phones[phone]++;//如果map中没有该key的话，value会自动初始化；如果有该key的话，则改变value值
                //此外，因为Map中的元素是自动按key升序排序,所以我就不要再排一次序啦，好省心！（也因为这样，所以不能对map用sort函数。）
	}
	bool hasDuplicate=false;
	for ( iter = phones.begin( ); iter != phones.end( ); iter++ ){
		int existNum=iter ->second;
		if(existNum>1){
			hasDuplicate=true;
			string pho=iter->first;
			for(int i=0;i<3;i++)
				cout<<pho[i];
			cout<<'-';
			for(int i=3;i<7;i++)
				cout<<pho[i];
            cout <<" "<<existNum<<endl;
		}
    } 
	if(hasDuplicate==false)
		cout<<"No duplicates."<<endl;
	return 0;
}

这样的代码（memory 5200k  time 1375ms）时间还是不行啊，究竟问题出在哪里了呢？我就去找别人的代码来研究。

被我找到一个(3476k 610ms)的：

#include <iostream>  
#include <cstdio>  
#include <cstring>  
#include <set>  
#include <map>  
  
using namespace std;  
  
int num[] =   
{  
    2, 2, 2,  
    3, 3, 3,  
    4, 4, 4,  
    5, 5, 5,  
    6, 6, 6,  
    7, 0, 7, 7,  
    8, 8, 8,  
    9, 9, 9  
};  
  
map<int, int> s;  
char buf[128];  
  
int main()  
{  
    int t;  
    scanf("%d", &t);  
    bool flag = false;  
    for(int i = 0; i < t; i++)  
    {  
        scanf("%s", &buf);  
        int c = 0;  
        for(int j = 0; buf[j]; j++)  
        {  
            if(isdigit(buf[j]))  
                c = c * 10 + buf[j] - '0';  
            else if(isalpha(buf[j]))  
                c = c * 10 + num[ buf[j] - 'A' ];  
        }  
        s[c]++;  
    }  
    for(map<int, int>::iterator it = s.begin(); it != s.end(); it++)  
        if(it->second > 1)  
        {  
            flag = true;  
            printf("%03d-%04d %d\n", it->first / 10000, it->first % 10000, it->second);  
        }  
    if(!flag)  
        puts("No duplicates.");  
return 0;  
}  

乍一看这人跟我的思路一样啊，代码也差不多。但是还是有区别的：

1.在map中，我将电话号码存成string，他是用int存储的，考虑到int的最大范围能有10个整数，所以存储7个整数是绰绰有余的。大概在map中，对于int的映射比对于string的映射要快吧。

2.在输出时，他用了printf("%03d-%04d %d\n", it->first / 10000, it->first % 10000, it->second); 这个技巧，输出7位整数的前3位和后4位，好巧妙啊！

我查了一下%03d这样的用法：

%3d--说明输出数据按三个长度的宽度显示，如果要输出的长度大于3时会忽略此时的域宽，以正常显示数据（就是把所有的数字都显示出来，域宽不起作用）。如果不足3位的话，会右对齐，在数据左边以空格补全的，使整个数据以三个长度的域宽显示。
%-3d--左对齐，右边填充空格。%-3d表示以整型十进制格式输出，宽度为3，负值表示左对齐，不足三位在右边补空格。
%03d---一种左边补0 的等宽格式,比如数字12,%03d出来就是: 012。 0(数字零)代表的是不足n位长度的左补齐0。

接下来看一个不用map的解法，使用了qsort。该解法 memory 644k, time 735ms。

//Memory Time
//644K  672MS 

/*Qsort*/

#include<iostream>
#include<algorithm>
#include<iomanip>
using namespace std;

void initial(int* ctoi)  //把字符ch转换为其在手机上对应的数字键
{
    for(int i=0;i<=9;i++)
        ctoi[i+'0']=i;

    ctoi['A'] = ctoi['B'] = ctoi['C'] = 2;
    ctoi['D'] = ctoi['E'] = ctoi['F'] = 3;
    ctoi['G'] = ctoi['H'] = ctoi['I'] = 4;
    ctoi['J'] = ctoi['K'] = ctoi['L'] = 5;
    ctoi['M'] = ctoi['N'] = ctoi['O'] = 6;
    ctoi['P'] = ctoi['R'] = ctoi['S'] = 7;
    ctoi['T'] = ctoi['U'] = ctoi['V'] = 8;
    ctoi['W'] = ctoi['X'] = ctoi['Y'] = 9;
    return;
}

int main(int i)
{
    int ctoi['Z'+1];
    initial(ctoi);

    int n;  //号码数
    while(cin>>n)
    {
        /*Initial*/

        int* sort_out=new int[n];  //按字典序存放待输出的电话号码

        /*Input*/

        for(i=0;i<n;i++)
        {
            int x=0;
            char s[20];
            cin>>s;

            for(int j=0;s[j]!='\0';j++)
            {
                if(s[j]=='-' || s[j]=='Q' || s[j]=='Z')
                    continue;
                x=x*10+ctoi[ s[j] ];
            }
            sort_out[i]=x;
        }

        /*Sort & Output*/

        sort(sort_out,sort_out+n);

        bool flag=true;  //标记是否所有号码都是唯一的
        for(i=0;i<n;)//注意哦，这里没有i++！因为下面已经把i增加了。
        {
            int time=0;  //ort_out[i]出现的次数
            int k=sort_out[i];
            bool sign=false;  //标记k出现次数是否大于2

            while(k==sort_out[i] && i<n)
            {
                time++;
                i++;

                if(time==2)
                {
                    flag=false;
                    sign=true;
                }
            }

            if(sign)
            {
                cout<<setfill('0')<<setw(3)<<k/10000;
                cout<<'-';
                cout<<setfill('0')<<setw(4)<<k%10000;
                cout<<' '<<time<<endl;
            }
        }
        if(flag)
            cout<<"No duplicates."<<endl;

        delete sort_out;
    }
    return 0;
}

他是先将所有的号码转成数字串，存在sort_out数组里，接着用#include<algorithm>里的sort(...，...)函数来排好序，之后在这个排好序的数组里找重复的元素。找的时候很巧妙，一遍就过掉了，发现重复的就打印。

研究了一下algorithm中的sort方法：

STL里面有个sort函数，可以直接对数组排序，复杂度为n*log2(n)。注意：缺省是升序排序。sort中一个改变排序顺序的例子如下（降序）：

#include<iostream>
#include<algorithm>
using namespace std;
bool cmp (const int a, const int b)
{
    return a > b;
}
int main()
{
    int data[5];
    for(int i = 0; i < 5; i++)
        cin >> data[i];
    sort(data, data + 5, cmp);
    return 0;
}

如需要对数组t的第0到len-1的元素排序，就写sort(t,t+len);对向量v排序也差不多，sort(v.begin(),v.end());排序的数据类型不局限于整数，只要是定义了小于运算的类型都可以，比如字符串类string。

接下来介绍一个典型的以空间换时间的解法。（memory:49296K , time:469ms)

#include<cstdio>  
#include<cstring>  
#include<string>  
#include<map>  
#include<set>  
#include<queue>  
#include<algorithm>  
using namespace std;  
int m[256];  
int vd[10000000];  
bool vs[10000000];  
priority_queue< int > q;  
int main()  
{     
    m['0']=0;                  m['1']=1;  
    m['2']=2;m['A']=2;m['B']=2;m['C']=2;  
    m['3']=3;m['D']=3;m['E']=3;m['F']=3;  
    m['4']=4;m['G']=4;m['H']=4;m['I']=4;  
    m['5']=5;m['J']=5;m['K']=5;m['L']=5;  
    m['6']=6;m['M']=6;m['N']=6;m['O']=6;  
    m['7']=7;m['P']=7;m['R']=7;m['S']=7;  
    m['8']=8;m['T']=8;m['U']=8;m['V']=8;  
    m['9']=9;m['W']=9;m['X']=9;m['Y']=9;  
    int n,len;  
    char ss[100],s[100];  
    scanf("%d",&n);  
    getchar();  
    for(int i=0;i<n;i++)  
    {  
        scanf("%s",s);  
        getchar();  
        len=strlen(s);        
        int num=0;  
        for(int j=0;j<len;j++)  
        {  
            if(s[j]=='-'||s[j]=='Q'||s[j]=='Z')  
                continue;  
            else  
            {  
                num=num*10+m[s[j]];  
            }  
        }  
        vd[num]++;  //vd记录出现次数
        if(!vs[num]&&vd[num]>=2)  //vs数组用来判断该数串有没有被存入q
            q.push(-num),vs[num]=1;  
    }  
    if(q.empty())  
        printf("No duplicates.\n");  
    while(q.empty()==0)  
    {  
        int tem=-q.top();  
        q.pop();  
        sprintf(ss,"%07d",tem);  
        for(int j=0;j<7;j++)  
            if(j==2)  
                printf("%c-",ss[j]);  
            else  
                printf("%c",ss[j]);  
        printf(" %d\n",vd[tem]);  
    }  
    return 0;  
}

看到vd[10000000];我就惊呆了，因为是7位整数，所以使用了10000000（后面7个0），将数组的index=号码数串，value放置出现次数。果然是以空间换时间啊，这样的话每次赋值到数组的某个index上，肯定比map要快啊！使用了优先级队列，存储重复的数串。因为普通的优先级队列按从大到小排序，所以作者用了一个小技巧，压入的是（-1）*真实的数串。然后拿出来的时候再*（-1）一遍，太机智了。

还用到了sprintf(),我又去查了一下：（c++烂伤不起啊T_T）

sprintf 最常见的应用之一莫过于把整数打印到字符串中，所以，spritnf 在大多数场合可以替代itoa。
如：
//把整数123 打印成一个字符串保存在s 中。
sprintf(s, "%d", 123); //产生"123"
可以指定宽度，不足的左边补空格：
sprintf(s, "%8d%8d", 123, 4567); //产生：" 123 4567"
当然也可以左对齐：
sprintf(s, "%-8d%8d", 123, 4567); //产生："123 4567"