poj 1256 Anagram
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Description
You are to write a program that has to generate all possible words from a given set of letters.
Example: Given the word "abc", your program should - by exploring all different combination of the three letters - output the words "abc", "acb", "bac", "bca", "cab" and "cba".
In the word taken from the input file, some letters may appear more than once. For a given word, your program should not produce the same word more than once, and the words should be output in alphabetically ascending order.
Example: Given the word "abc", your program should - by exploring all different combination of the three letters - output the words "abc", "acb", "bac", "bca", "cab" and "cba".
In the word taken from the input file, some letters may appear more than once. For a given word, your program should not produce the same word more than once, and the words should be output in alphabetically ascending order.
Input
The input consists of several words. The first line contains a number giving the number of words to follow. Each following line contains one word. A word consists of uppercase or lowercase letters from A to Z. Uppercase and lowercase letters are to be considered different. The length of each word is less than 13.
Output
For each word in the input, the output should contain all different words that can be generated with the letters of the given word. The words generated from the same input word should be output in alphabetically ascending order. An upper case letter goes before the corresponding lower case letter.
Sample Input
3aAbabcacba
Sample Output
AabAbaaAbabAbAabaAabcacbbacbcacabcbaaabcaacbabacabcaacabacbabaacbacabcaacaabcabacbaa
Hint
An upper case letter goes before the corresponding lower case letter.
So the right order of letters is 'A'<'a'<'B'<'b'<...<'Z'<'z'.
So the right order of letters is 'A'<'a'<'B'<'b'<...<'Z'<'z'.
题意:给出一字符串,求出所有按字母顺序排序的全排列。
#include <iostream>#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;char num[15],temp[15];int n;bool vis[15];bool cmp(const char &a, const char &b){ int x = a - 'A'; int y = b - 'A'; if (x >= 26) { x -= 'a' - 'A'; } if (y >= 26) { y -= 'a' - 'A'; } if (x != y) { return x < y; } return a < b;}void solve(int cnt){ bool flag[100]={0}; //用于去重, if(cnt==n) { temp[n]='\0'; printf("%s\n",temp); return; } for(int i=0;i<n;i++) { if(!vis[i] && flag[num[i]-'A']==0) //flag 递归到该层中num[i]没有枚举过 { vis[i]=true; temp[cnt]=num[i]; flag[num[i]-'A']=1; solve(cnt+1); vis[i]=false; } }}int main(){ int t; scanf("%d",&t); while (t--) { scanf("%s",num); n=strlen(num); sort(num,num+n, cmp); memset(vis, false, sizeof(vis)); solve(0); } return 0;}
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