POJ 字符串的全排列 水题也搞我 多种方法 DFS，STL，直接模拟。。。。。。

Orders

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 8100 Accepted: 5048

Description

The stores manager has sorted all kinds of goods in an alphabetical order of their labels. All the kinds having labels starting with the same letter are stored in the same warehouse (i.e. in the same building) labelled with this letter. During the day the stores manager receives and books the orders of goods which are to be delivered from the store. Each order requires only one kind of goods. The stores manager processes the requests in the order of their booking.

You know in advance all the orders which will have to be processed by the stores manager today, but you do not know their booking order. Compute all possible ways of the visits of warehouses for the stores manager to settle all the demands piece after piece during the day.

Input

Input contains a single line with all labels of the requested goods (in random order). Each kind of goods is represented by the starting letter of its label. Only small letters of the English alphabet are used. The number of orders doesn't exceed 200.

Output

Output will contain all possible orderings in which the stores manager may visit his warehouses. Every warehouse is represented by a single small letter of the English alphabet -- the starting letter of the label of the goods. Each ordering of warehouses is written in the output file only once on a separate line and all the lines containing orderings have to be sorted in an alphabetical order (see the example). No output will exceed 2 megabytes.

Sample Input

bbjd

Sample Output

bbdjbbjdbdbjbdjbbjbdbjdbdbbjdbjbdjbbjbbdjbdbjdbb

Source

题意：

输入字符串  输出这个串的所有派生串   木有重复的哦     字符串长度最多200     都是字母小写

我这样写  直接超时了   很是郁闷  放进map里怎么会超时那  我的想法是放进map可以防止重复   其实next_permutation 根本就不会产生重复串

他产生的是字典序改变的串（这个是字典序上升）

#include<stdio.h>#include<string.h>#include<string>#include<algorithm>#include<iostream>#include<map>using namespace std;int main(){char s[300];int i,cnt;map<string,int>mp;map<string,int>::iterator it;while(scanf("%s",s)!=EOF){mp.clear();sort(s,s+strlen(s));//puts(s);cnt=strlen(s);do{        mp[s]=1;}while(next_permutation(s,s+cnt));for(it=mp.begin();it!=mp.end();it++)cout<<it->first<<endl;//注意这里哦    不要用printf  不支持 会有弹窗}return 0;}

后来改成这样就过了  32ms

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;int main(){char s[300];int i,cnt;while(scanf("%s",s)!=EOF){sort(s,s+strlen(s));cnt=strlen(s);do{           puts(s); }while(next_permutation(s,s+cnt));}return 0;}

下面是DFS方法  并且去重

/*去重的方法是，如果当前的a的flag为1，即被使用了，而前一个a未被使用，则跳出。用程序表示为flag[i]==1&&flag[i-1]==0&&s[i]==s[i-1]&&i>0     s中存放的是排序后的字符串*/#include <stdio.h>#include <string.h>#include <stdlib.h>#include<algorithm>using namespace std;int len;int flag[201];char str[201];char res[201];void dpoutput(int n){int i;if (n==len){printf("%s\n", res);return;}for (i=0;i<len;i++){if (flag[i]||(i>0&&flag[i-1]==0&&str[i]==str[i-1]))continue;res[n]=str[i];flag[i] = 1;dpoutput(n+1);flag[i]=0;}}int main(){while (scanf("%s", str) != EOF){len = strlen(str);memset(flag, 0, sizeof(flag));sort(str,str+len);dpoutput(0);}return 0;}

直接模拟   明天写。。。。。。