紫书 例题7-5 UVA 129

【题意】输入正整数n，l，输出由前l个字符组成的，字典序第k小的没有两个相邻重复子串的串。

【解题方法】判断ABCABA是否包含重复子串的方法是这样的：运用回溯法，只要比较当前串后缀，因为回溯法的特性，我们只要考虑当前的情况即可，前面的不用管，这样递归可以保证从头到尾都是成立的，当然后缀要从一个比较到n+1/2个（奇数多考虑一个），例如：ABCABA，先比较A与前面的B，再比较BA与CA，以此类推。

【AC代码】

////Created by just_sort 2016/1/5//Copyright (c) 2016 just_sort.All Rights Reserved////#include <ext/pb_ds/assoc_container.hpp>//#include <ext/pb_ds/tree_policy.hpp>//#include <ext/pb_ds/hash_policy.hpp>#include <set>#include <map>#include <queue>#include <stack>#include <cmath>#include <cstdio>#include <time.h>#include <cstdlib>#include <cstring>#include <sstream> //isstringstream#include <iostream>#include <algorithm>using namespace std;//using namespace __gnu_pbds;typedef long long LL;typedef pair<int, LL> pp;#define REP1(i, a, b) for(int i = a; i < b; i++)#define REP2(i, a, b) for(int i = a; i <= b; i++)#define REP3(i, a, b) for(int i = a; i >= b; i--)#define CLR(a, b)     memset(a, b, sizeof(a))#define MP(x, y)      make_pair(x,y)const int maxn = 1e3 + 10;const int maxm = 2e5;const int maxs = 10;const int maxp = 1e3 + 10;//const int INF  = 1e9;const int UNF  = -1e9;const int mod  = 1e9 + 7;int gcd(int x, int y) {return y == 0 ? x : gcd(y, x % y);}//typedef tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update>order_set;//headint n, L, cnt, qq;int s[100];int dfs(int cur){    if(cnt++ == n){        for(int i = 0; i < cur; i++){            qq++;            printf("%c", 'A' + s[i]);            if(qq % 64 == 0) printf("\n");            else if(qq % 4 == 0 && i != cur - 1) printf(" ");        }        printf("\n%d\n", qq);        return 0;    }    for(int i = 0; i < L; i++){        s[cur] = i;        int ok = 1;        for(int j = 1; j * 2 <= cur + 1; j++){            int ok2 = 1;            for(int k = 0; k < j; k++){                if(s[cur - k] != s[cur - k - j]){                    ok2 = 0;                    break;                }            }            if(ok2){                ok = 0;                break;            }        }        if(ok){            if(!dfs(cur + 1)) return 0;        }    }    return 1;}int main(){    while(scanf("%d%d", &n, &L) != EOF)    {        if(n == 0 && L == 0) break;        memset(s, 0, sizeof(s));        qq = cnt = 0;        dfs(0);    }    return 0;}