CDOJ 1305 Just a Magic String

子串只有10^6，所以可以直接去匹配，不管什么规律，但是，母串要到4*10^6，才能考虑到10^6的串的所有情况，具体，用小数据列举一下就可以发现。

然后我用的是KMP，貌似暴力匹配也能过，另，正解貌似是DFS

代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
using namespace std;
#define maxn 4194304
char s1[maxn], s2[1000003];
int then[1000003];
char idx[256];
int n1, n2;
void init_s1()
{
idx['a'] = 'b', idx['b'] = 'a';
s1[0] = 'a', s1[1] = 'b';
int len = 2;
while (len < maxn)
{
for (int i = 0; i < len; ++i)
s1[len + i] = idx[s1[i]];

len <<= 1;
}
}
void get_then()
{
int i = 0, j = -1;
then[0] = -1;
while (i < n2)
{
if (j == -1 || s2[i] == s2[j])
{
++i;
++j;
then[i] = j;
}
else
j = then[j];
}
}
int kmp()
{
int t1 = 0, t2 = 0;
while (t1<n1&&t2<n2)
{
if (t2 == -1 || s1[t1] == s2[t2])
{
t1++;
t2++;
}
else t2 = then[t2];
}
if (t2 == n2)
return t1 - t2 + 1;
else
return -1;
}
int main()
{
//freopen("input.txt", "r", stdin);
scanf("%s", s2);
n1 = maxn, n2 = strlen(s2);
init_s1();
get_then();
/*for (int i = 0; i < 10; ++i)
printf("%c", s1[i]);*/
int ans = kmp();
printf("%d\n", ans);
//system("pause");
//while (1);
return 0;
}