poj 1002
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研究了半天，结果果然是超时了。其实数据处理的时间应该不长，关键是读入数据的时间太长了。还是记下来留作念想吧。
#include "poj.h"#include <iostream>#include <string>#include <vector>#include <algorithm>using namespace std;char Codes[31] = {'0','0','A','D','G','J','M','P','T','W', '0','0','B','E','H','K','N','R','U','X', '0','0','C','F','I','L','O','S','V','Y'};class Word{public:string w;int count;bool operator ==(const Word &value) {return value.w == this->w; }bool operator>(const Word& value){return this->w > value.w;}bool operator<(const Word& value){return this->w < value.w;}};string translate(string original){string result = "";for(int i = 0; i < original.length();i++){char c = original[i];if(c == '-')continue;if(c >= '0' && c<= '9'){result += c;}else{//find the number c representsfor(int j = 0; j < 30;j++){if(c == Codes[j]){result += j%10 + '0';break;}}}}return result;}void p1002(){int i,n;cin>>n;vector<string> data;vector<Word> dumpnumber;for(i = 0; i < n; i++){string a;cin>>a;string t = translate(a);if( find(data.begin(), data.end(), t) != data.end()){Word word;word.w = t;vector<Word>::iterator it = find(dumpnumber.begin(), dumpnumber.end(),word);if(it != dumpnumber.end()){it->count++;}else{word.count=2;dumpnumber.push_back(word);}}data.push_back(t);}sort(dumpnumber.begin(),dumpnumber.end());vector<Word>::iterator it = dumpnumber.begin();if(dumpnumber.empty()){cout<<"No duplicates."<<endl;}while(it != dumpnumber.end()){for(i = 0; i < 3; i ++)cout<<it->w[i];cout<<"-";for(;i<7;i++)cout<<it->w[i];cout<<" "<<it->count<<endl;it++;}}终于决定用暴力方法，写了个丑陋的办法，果然AC了，用时1200MS：
#include <iostream>#include <string>#include <sstream>//#include "poj.h"#using <System.dll>using namespace System;using namespace System::Diagnostics;using namespace std;char Codes[31] = {'0','0','A','D','G','J','M','P','T','W', '0','0','B','E','H','K','N','R','U','X', '0','0','C','F','I','L','O','S','V','Y'};int translate(string original){int result = 0, index = 1000000;for(int i = 0; i < original.length();i++){char c = original[i];if(c == '-')continue;if(c >= '0' && c<= '9'){result += index * (c - '0');index /= 10;}else{//find the number c representsfor(int j = 0; j < 30;j++){if(c == Codes[j]){result += index * (j%10);index /= 10;break;}}}}return result;}int count[10000000] = {0};int main(){int i,t,n,flag = -1;string a;stringstream ss;cin>>n;Stopwatch ^ stopWatch;stopWatch = Stopwatch::StartNew();for(i = 0; i < n; i++){cin >> a;t = translate(a);count[t]++;}for(i = 0; i < 10000000; i++){int c = count[i];if(c<2)continue;if(flag<0)flag = 1;ss.clear();ss<<i;ss>>a;while(a.length() < 7) a = '0' + a;cout<<a[0]<<a[1]<<a[2]<<"-"<<a[3]<<a[4]<<a[5]<<a[6]<<" "<<c<<endl;}if(flag<0)cout<<"No duplicates."<<endl;stopWatch->Stop();Int64 ticksThisTime = stopWatch->ElapsedMilliseconds;cout<<"time:"<<ticksThisTime<<endl;system("PAUSE");return 0;}同时学学别人家的代码：
先排序再找重复？还是找出重复再排序？这是个问题。
我试着换成这种输入方式，但并没有让我的代码更快，反而TLE了。不过多重IF的语句确实比for循环去char数组里找快一些貌似。
但是C语言天然优势使得这段代码只用了800MS左右。
#include <stdio.h>#include <string.h>#include <stdlib.h>#include <ctype.h>char pn[100000][8];void main (){ int i, j, n, c; int none = 1; scanf ("%d", &n); for (i = 0; i < n; i++) { j = 0; while (j < 7){ if (!isalnum (c = getchar ())) continue; if (c == 'A' || c == 'B' || c == 'C') c = '2'; if (c == 'D' || c == 'E' || c == 'F') c = '3'; if (c == 'G' || c == 'H' || c == 'I') c = '4'; if (c == 'J' || c == 'K' || c == 'L') c = '5'; if (c == 'M' || c == 'N' || c == 'O') c = '6'; if (c == 'P' || c == 'R' || c == 'S') c = '7'; if (c == 'T' || c == 'U' || c == 'V') c = '8'; if (c == 'W' || c == 'X' || c == 'Y') c = '9'; pn[i][j++] = c;} pn[i][7] = '\0'; } qsort (pn, n, 8, (int (*) (const void *, const void *)) strcmp); j = 0; for (i = 1; i < n; i++) { if (strcmp (pn[i], pn[j]) != 0) {if (i - j > 1) {printf ( "%c%c%c-%s %d\n", pn[j][0], pn[j][1], pn[j][2], pn[j] + 3, i - j);none = 0; }j = i; } } if (i - j > 1) { printf ("%c%c%c-%s %d\n", pn[j][0], pn[j][1], pn[j][2], pn[j] + 3, i - j ); none = 0; } if (none) printf ("No duplicates.\n"); system("PAUSE");}这是另一段代码，思路也是先排序，然后一边找重复一边输出：(转自http://blog.chinaunix.net/uid-8059407-id-2034320.html)
#include <stdio.h>#include <stdlib.h>#include <string.h>char map[26] = {'2', '2', '2', '3', '3', '3', '4', '4', '4', '5', '5', '5', '6', '6', '6', '7', '7', '7', '7', '8', '8', '8', '9', '9', '9', '9'};//按键映射表char phone[100000][9]; //全部电话号码int n; //电话号码的数量char number[80]; //一个未经处理的电话号码int main(void){ int loop; //循环变量 int phone_i=0, phone_j=0; //phone数组的下标 int tag = 0; //重复标志，0没有处于一个重复，1有重复 int count = 1; //重复次数 scanf("%d", &n); for(phone_i=0; phone_i<n; phone_i++){//格式化所有输入数据为111-1111的形式 scanf("%s", number); for(loop=0; number[loop]!='\0'; loop++){//检查每一个字符 if(phone_j == 3){//第四个字符必须是'-' phone[phone_i][phone_j] = '-'; phone_j++; } if(number[loop]>='0' && number[loop]<='9'){//当前字符是数字 phone[phone_i][phone_j] = number[loop]; } else if(number[loop]>='A' && number[loop]<='Z'){//当前字符是大写字母 phone[phone_i][phone_j] = map[number[loop]-'A']; } else if(number[loop] == '-'){//忽略'-' continue; } phone_j++; } phone_j = 0; } //排序全部电话号码 qsort(phone, n, 9, strcmp); for(loop=1; loop<n; loop++){//检查重复次数，输出结果 if(strcmp(phone[loop], phone[loop-1]) == 0){ tag = 1; count++; } else{//没有重复 if(count > 1){ printf("%s %d\n", phone[loop-1], count); } count = 1; } } if(count > 1){//查看最后一条记录是否有重复 printf("%s %d\n", phone[loop-1], count); } if(!tag) printf("No duplicates.\n"); return 0;}这个很有借鉴意义，根据输入来确定自定义数据指针的长度：（转自http://blog.csdn.net/hello_cosmos/article/details/8683599）
#include <stdio.h>#include <stdlib.h>#include <string.h>typedef struct __data{ int telNum; int repTim;} da;int main(){da *data;int numOfRecord,i;char temp[2000]; //数组开小了会WA。。。我了个擦。。。多么变态的测试数据啊。。第二第三次WA都是这个问题scanf("%d",&numOfRecord);data=(da *)calloc(numOfRecord,sizeof(da));for (i=0;i<numOfRecord;i++){scanf("%s",temp);stdlizeNum(temp,&(data[i]));}sortPrintf(data,numOfRecord); free(data); return 0;}int stdlizeNum(char *num,da *ans){ int len=strlen(num); int i; int k=0; ans->repTim=0; ans->telNum=0; for (i=0;i<len;i++){ switch(num[i]){ case 'A': case 'B': case 'C': ans->telNum+=2; ans->telNum*=10; break; case 'D': case 'E': case 'F': ans->telNum+=3; ans->telNum*=10; break; case 'G': case 'H': case 'I': ans->telNum+=4; ans->telNum*=10; break; case 'J': case 'K': case 'L': ans->telNum+=5; ans->telNum*=10; break; case 'M': case 'N': case 'O': ans->telNum+=6; ans->telNum*=10; break; case 'P': case 'R': case 'S': ans->telNum+=7; ans->telNum*=10; break; case 'T': case 'U': case 'V': ans->telNum+=8; ans->telNum*=10; break; case 'W': case 'X': case 'Y': ans->telNum+=9; ans->telNum*=10; break; case '-': break; default: ans->telNum+=(int)(num[i]-'0'); ans->telNum*=10; break; } } ans->telNum/=10;}int sortCmp_fun(const void *a,const void *b){ return ((*(da *)a).telNum)-((*(da *)b).telNum);}int sortPrintf(da *input,int len){int i,offOn;qsort(input,len,sizeof(da),sortCmp_fun); for (i=0;i<len;i++){ if (input[i].telNum==input[i-1].telNum){ input[i].repTim=input[i-1].repTim+1; input[i-1].repTim=0; } } offOn=0; for (i=0;i<len;i++){ if (input[i].repTim!=0){ offOn=1; printf("%03d-%04d %d\n",input[i].telNum/10000,input[i].telNum%10000,input[i].repTim+1); //"%d"这里不会出'0',而"%0d"会打印出来，第一次WA就是这里出错了 } } if (offOn==0){ printf("No duplicates.\n"); }}#include <stdio.h>#include <string.h>#include <stdlib.h>#define UNCODE(a) (a>='A'?((a>'Q'?(a-'A'-1):(a-'A'))/3+2):a-'0')//#define DEBUGint uncode(char* s){int i;int result=0;for(i=0;i<strlen(s);i++){if(s[i]=='-') continue;result=result*10+UNCODE(s[i]);}return result;}int cmp(const void* a,const void* b){return *(int*)a-*(int*)b;}int main(){#ifdef DEBUGfreopen("data.dat","r",stdin);//freopen("out.dat","w",stdout);#endifint* store;char s[50];int time;int n=0;int i;int start,value;int dup=0;scanf("%d",&time);store=(int*)malloc(sizeof(int)*(time));for(n=0;n<time;n++){scanf("%s",s);store[n]=uncode(s);} qsort(store,time,sizeof(store[0]),cmp);value=store[0];start=0;for(n=0;n<time;n++){if(store[n]!=value){ if((n-start)>1){printf("%03d-%04d %d\n",store[start]/10000,store[start]%10000,n-start);dup=1;}start=n;value=store[n];}}if((n-start)>1){printf("%03d-%04d %d\n",store[start]/10000,store[start]%10000,n-start);dup=1;}if(!dup){printf("No duplicates.\n");}free(store);return 1;}/* Copyright Derek Kisman (ACM ICPC ECNA 1999) */char map[26] = {'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', 0};char ph[100000][9];int nph;char buf[1000];main() {int i, j, k, x, y, z, n;char ch;memset( ph, 0, sizeof(ph) );scanf( " %d", &nph );for( i = 0; i < nph; i++ ) {scanf( " %s", buf );x = 0;for( j = 0; buf[j]; j++ ) {if( buf[j] == '-' ) continue;if( buf[j] >= 'A' && buf[j] <= 'Z' ) buf[j] = map[buf[j]-'A'];ph[i][x++] = buf[j];if( x == 3 ) ph[i][x++] = '-';}}qsort( ph, nph, 9, strcmp );x = 1; z = 0;for( i = 1; i < nph; i++ ) {if( strcmp( ph[i-1], ph[i] ) ) {if( x > 1 ) {printf( "%s %d\n", ph[i-1], x );z = 1;}x = 1;} else {x++;}}if( x > 1 ) {printf( "%s %d\n", ph[i-1], x );z = 1;}if( !z ) printf( "No duplicates.\n" );} 0 0
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