SGU142 Keyword

SGU142 Keyword

题目大意

给出一个”ab”构成的字符串，找出最短的”ab”串不是它的子串

算法思路

长度为20的串有2202^{20}种，超过N的长度

因此我们可以枚举串长，然后遍历原串找出是否有没出现过的串

这里利用了位运算与bitset加速这一过程

时间复杂度: O(NlogN)O(NlogN)

代码

/**
* Copyright © 2015 Authors. All rights reserved.
*
* FileName: 142.cpp
* Author: Beiyu Li <sysulby@gmail.com>
* Date: 2015-06-17
*/
#include <bits/stdc++.h>

using namespace std;

#define rep(i,n) for (int i = 0; i < (n); ++i)
#define For(i,s,t) for (int i = (s); i <= (t); ++i)
#define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)

typedef long long LL;
typedef pair<int, int> Pii;

const int inf = 0x3f3f3f3f;
const LL infLL = 0x3f3f3f3f3f3f3f3fLL;

const int maxn = 500000 + 5;
const int maxs = (1 << 20) + 5;

int n;
char s[maxn];

int main()
{
scanf("%d%s", &n, s);
For(i,1,n) {
bitset<maxs> vis;
int mask = (1 << i) - 1, S = 0;
rep(j,i-1) S = S << 1 | (s[j] - 'a');
for (int j = i - 1; j < n; ++j) {
S = (S << 1 | (s[j] - 'a')) & mask;
vis.set(S);
}
if (vis.count() < (1 << i)) {
printf("%d\n", i);
For(S,0,mask) if (!vis.test(S)) {
for (int j = i - 1; j >= 0; --j)
putchar((1 << j) & S? 'b': 'a');
puts("");
break;
}
break;
}
}

return 0;
}