lightoj-1023 - Discovering Permutations(全排列)

1023 - Discovering Permutations
PDF (English) Statistics Forum
Time Limit: 0.5 second(s) Memory Limit: 32 MB
In this problem you have to find the permutations using the first N English capital letters. Since there can be many permutations, you have to print the first K permutations.

Input
Input starts with an integer T (≤ 100), denoting the number of test cases.

Each case contains two integers N, K (1 ≤ N ≤ 26, 1 ≤ K ≤ 30).

Output
For each case, print the case number in a line. Then print the first K permutations that contain the first N English capital letters in alphabetical order. If there are less than K permutations then print all of them.

Sample Input
Output for Sample Input
2
3 8
10 10
Case 1:
ABC
ACB
BAC
BCA
CAB
CBA
Case 2:
ABCDEFGHIJ
ABCDEFGHJI
ABCDEFGIHJ
ABCDEFGIJH
ABCDEFGJHI
ABCDEFGJIH
ABCDEFHGIJ
ABCDEFHGJI
ABCDEFHIGJ
ABCDEFHIJG

 

解题思路: 全排列模板改进一下，直接过。

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<string>
using namespace std;

char a[26]={'A','B','C','D','E','F','G',
            'H','I','J','K','L','M','N',
            'O','P','Q','R','S','T','U',
            'V','W','X','Y','Z'};
int T,n,k,cnt;
string str[130];

bool permutation(int num){
    if(num>=n){
        int i;int sum;
        sum = 0;
            str[cnt] = a;    
        str[cnt][n] = '\0';
//        cout<<str[cnt]<<endl;
        cnt++;
        if(cnt>=120) return false;
        return true;
    }
    for(int i=num;i<n;i++){
        swap(a[num],a[i]);
        if(!permutation(num+1)) return false; //这里直接返回后没执行下面的swap, 所以下一个case的时候要先把a[]重新排好序。 
        swap(a[num],a[i]);
    }
    return true;
}



int main(){
    
    
    scanf("%d",&T);
    for(int t=1;t<=T;t++){
        cnt = 0;
        scanf("%d%d",&n,&k);
        printf("Case %d:\n",t);

        sort(a,a+26);
        permutation(0);
        sort(str,str+cnt);
        for(int i=0;i<min(cnt,k);i++) {
            for(int j=0;j<n;j++) cout<<str[i][j];    
            cout<<endl;
        }
// 用全排列的STL的写法。    
//        do{
//            for(int i=0;i<n;i++) printf("%c",a[i]);
//            printf("\n");
//            cnt++;    
//        }while(next_permutation(a,a+n)&&(k>cnt));
        
    }
    return 0;
}