CodeForces 748B Santa Claus and Keyboard Check

题意：给定两个字符串a和b，问有多少种不同的字母组合对，使得将这些字母对替换字符串b后，可以变成字符串a。注意字母对彼此各不相同。

分析：vis[u]记录与u可形成关系的字母，若u与v不同，则形成字母对。若之后该关系被打破，则输出-1。

#pragma comment(linker, "/STACK:102400000, 102400000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define Min(a, b) ((a < b) ? a : b)
#define Max(a, b) ((a < b) ? b : a)
const double eps = 1e-8;
inline int dcmp(double a, double b){
    if(fabs(a - b) < eps) return 0;
    return a > b ? 1 : -1;
}
typedef long long LL;
typedef unsigned long long ULL;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const int MAXN = 1000 + 10;
const int MAXT = 10000 + 10;
using namespace std;
char a[MAXN], b[MAXN];
const string s = "abcdefghijklmnopqrstuvwxyz";
map<char, int> mp;
int vis[30];
int ans[30];
void init(){
    for(int i = 0; i < 26; ++i){
        mp[s[i]] = i + 1;
    }
}
int main(){
    init();
    while(scanf("%s%s", a, b) == 2){
        memset(vis, 0, sizeof vis);
        memset(ans, 0, sizeof ans);
        int len = strlen(a);
        int cnt = 0;
        bool ok = true;
        for(int i = 0; i < len; ++i){
            int tmpa = mp[a[i]], tmpb = mp[b[i]];
            if(!vis[tmpa] && !vis[tmpb]){
                vis[tmpa] = tmpb;
                vis[tmpb] = tmpa;
                if(tmpa != tmpb){
                    ans[cnt++] = tmpa;
                }
            }
            else if(vis[tmpa] != tmpb || vis[tmpb] != tmpa){
                ok = false;
                break;
            }
        }
        if(!ok){
            printf("-1\n");
        }
        else{
            printf("%d\n", cnt);
            for(int i = 0; i < cnt; ++i){
                printf("%c %c\n", ans[i] + 'a' - 1, vis[ans[i]] + 'a' - 1);
            }
        }
    }
    return 0;
}