UVA - 156 Ananagrams

题目大意： 输出不能重排成另外一个单词的单词，在比较时不分大小写，但是输出时要分。

map的基本应用

Input

Input will consist of a series of lines. No line will be more than 80 characters long, but may contain any

number of words. Words consist of up to 20 upper and/or lower case letters, and will not be broken

across lines. Spaces may appear freely around words, and at least one space separates multiple words

on the same line. Note that words that contain the same letters but of differing case are considered to

be anagrams of each other, thus ‘tIeD’ and ‘EdiT’ are anagrams. The file will be terminated by a line

consisting of a single ‘#’.

Output

Output will consist of a series of lines. Each line will consist of a single word that is a relative ananagram

in the input dictionary. Words must be output in lexicographic (case-sensitive) order. There will always

be at least one relative ananagram.

Sample Input

ladder came tape soon leader acme RIDE lone Dreis peat

ScAlE orb eye Rides dealer NotE derail LaCeS drIed

noel dire Disk mace Rob dries

#

Sample Output

Disk

NotE

derail

drIed

eye

ladder

soon

#include <cstdio>
#include <iostream>
#include <map>
#include <cstring>
#include <algorithm>
#include <set>
using namespace std;
#define maxn 1005
map<string,string> data;
map<string,int>vis;

char m[maxn],s[maxn];
int main()
{
char c;
int flag = 0;
int num = 0;
int sss = 0;
while(c = getchar())
{
if(c == EOF || c=='#')break;
if((c <= 'z' && c>='a') || (c>='A' && c<='Z')) s[num++] = c;
if(!((c <= 'z' && c>='a') || (c>='A' && c<='Z')) && num != 0)
{
s[num] = '\0';
strcpy(m,s);
m[num] = '\0';
for(int i = 0; i < num ;i ++)
if(isupper(m[i]))
m[i] = m[i]+('a' - 'A');
sort(m,m+num);
int flag = 0;
data[s] = m;
if(vis.count(m))
vis[m]++;
else
vis[m] = 1;

num = 0;
}
}

map<string,string>::iterator it = data.begin();
for(; it != data.end(); it++)
{
if(vis[it->second] == 1)
printf("%s\n",it->first.c_str());
}

}