pku 1731 Orders

简单的不重复排列问题

//法1 让脚标i自加消重复
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <iostream>

using namespace std;

#define MAX 205

char str[MAX],out[MAX];
int len;
bool used[MAX];

void dfs(int cur)
{
    if(cur==len)
    {
        printf("%s\n",out);
    }
    else
        for(int i=0; i<len; i++)
        {
            if(!used[i])
            {
                out[cur]=str[i];
                used[i]=true;
                dfs(cur+1);
                used[i]=false;

                i++;
                while( i<len && str[i]==str[i-1])
                    i++;
                i--;
            }
        }
}

int main()
{
    while(scanf("%s",str)!=EOF)
    {
        len=strlen(str);
        memset(used,false,sizeof(used));
        sort(str,str+len);
        dfs(0);
    }
    return 0;
}

//****************************************************

 

//法2 某牛的hash做法

#include <stdio.h>
#include <string.h>
#include <algorithm>

using namespace std;

#define SIZE 100
#define HASHSIZE 26

int len;
char str[SIZE];
char out[SIZE];
int hash[HASHSIZE];

void DFS(int cur)
{
    if(cur == len)
    {
        out[cur] = 0;
        printf("%s\n", out);
    }
    else
    {
        for(int i = 0; i < HASHSIZE; i++)
        {
            if(hash[i] > 0)
            {
                out[cur] = 'a' + i;
                hash[i] --;
                DFS(cur + 1);
                hash[i] ++;
            }
        }
    }
}

int main()
{
    while(scanf("%s", str) != EOF)
    {
        getchar();
        memset(hash, 0, sizeof(hash));
        int i;
        len = strlen(str);
        for(i = 0; i < len; i++)
        {
            hash[str[i] - 'a'] ++;
        }
        DFS(0);
    }
    return 0;
}

把一堆相同的字母都放在一个相同的hash[]里，当for(int i = 0; i < HASHSIZE; i++)这句i自加是，就跳过了这堆相同的字母了，比第一种方法更巧，更快...