字符串处理 Codeforces Round #296 (Div. 2) B. Error Correct System

 /*
无算法
三种可能：1.交换一对后正好都相同，此时-2
2.上面的情况不可能，交换一对后只有一个相同，此时-1
3.以上都不符合，则不交换，-1 -1

*/
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <string>
#include <map>
#include <set>
#include <vector>
#include <set>
using namespace std;

const int MAXN = 2e5 + 10;
const int INF = 0x3f3f3f3f;

char s[MAXN], t[MAXN];
int p[30][30];

bool check1(int cnt)
{
for (int i=0; i<26; ++i)
{
for (int j=0; j<26; ++j)
{
if (p[i][j] && p[j][i])
{
printf ("%d\n", cnt-2);
printf ("%d %d\n", p[i][j], p[j][i]);
return true;
}
}
}

return false;
}

bool check2(int cnt)
{
for (int i=0; i<26; ++i)
{
for (int j=0; j<26; ++j)
{
if (p[i][j])
{
for (int k=0; k<26; ++k)
{
if (p[k][i])
{
printf ("%d\n", cnt-1);
printf ("%d %d\n", p[i][j], p[k][i]);
return true;
}
}
}
}
}

return false;
}

void work(int n)
{
int cnt = 0;
for (int i=0; i<n; ++i)
{
if (s[i] != t[i])
{
p[s[i]-'a'][t[i]-'a'] = i + 1;
cnt++;
}
}
if (check1 (cnt))   return ;
if (check2 (cnt)) return ;

printf ("%d\n", cnt);
puts ("-1 -1");
}

int main(void)
{
//freopen ("B.in", "r", stdin);

int n;
while (~scanf ("%d", &n))
{
memset (p, 0, sizeof (p));
scanf ("%s", &s);
scanf ("%s", &t);

work (n);
}

return 0;
}