codeforces 585 div2C

传送门

题意：
给出长度为$n$的两个字符串，构造一个次数最少的交换二元组$（x,y）$，表示交换$s1[x]$和$s2[y]$，最后使得两个字符串相同

思路：
首先相同的字符就没必要去交换了。考虑上下字符不同的情况。上$a$,下$b$(下文称$ab,ba$同理)
对于$ab$如果有偶数组$ab$,我们可以进行$nu{m}_{ab}/2$次交换就可以使得上下相同，$ba$同理。

如果有奇数组，那么我们可以先交换一次，将它们变成偶数组，那么就是暴力两两匹配交换。

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<stdio.h>
#include<string.h>
#include<queue>
#include<cmath>
#include<map>
#include<set>
#include<vector> 
using namespace std;
#define inf 0x3f3f3f3f
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define mem(a,b) memset(a,b,sizeof(a));
#define lowbit(x)  x&-x;  
#define debugint(name,x) printf("%s: %d\n",name,x);
#define debugstring(name,x) printf("%s: %s\n",name,x);
typedef long long ll;
typedef unsigned long long ull;
const double eps = 1e-6;
const int maxn = 2e5+5;
const int mod = 1e9+7;
inline int read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
char s1[maxn],s2[maxn];
int n;
int ans1[maxn], ans2[maxn];
int main() {
    scanf("%d",&n);
    scanf("%s%s",s1+1,s2+1);
    int ab = 0, ba = 0;
    for(int i = 1; i <= n; i++){
        if(s1[i] == s2[i]) continue;
        if(s1[i] == 'a' && s2[i] == 'b'){
            ab++;
        }else if(s1[i] == 'b' && s2[i] == 'a'){
            ba++;
        }
    }
    int cnt = 0;
    for(int i = 1; i <= n; i++){
        if(s1[i] == s2[i]) continue;
        if(s1[i] == 'a' && s2[i] == 'b'){
            if(ab&1){
                ans1[cnt] = i;
                ans2[cnt++] = i;
                s1[i] = 'b';
                s2[i] = 'a';
                break;
            }
        }else if(s1[i] == 'b' && s2[i] == 'a'){
            if(ba&1){
                ans1[cnt] = i;
                ans2[cnt++] = i;
                s1[i] = 'a';
                s2[i] = 'b';
                break;
            }
        }
    }
    int p1 = 0, p2 = 0;
    for(int i = 1; i <= n; i++){
        if(s1[i] == s2[i]) continue;
        if(s1[i] == 'a' && s2[i] == 'b'){
            if(!p1) p1 = i;
            else{
                s1[p1] = 'b';
                s2[i] = 'a';
                ans1[cnt] = p1;
                ans2[cnt++] = i;
                p1 = 0;

            }
        }else if(s1[i] == 'b' && s2[i] == 'a'){
            if(!p2) p2 = i;
            else{
                s1[p2] = 'a';
                s2[i] = 'b';
                ans1[cnt] = p2;
                ans2[cnt++] = i;
                p2 = 0;
            }
        }
    }

    if(strcmp(s1+1,s2+1)!=0) puts("-1");
    else{
        printf("%d\n",cnt);
        for(int i = 0; i < cnt; i++)
        printf("%d %d\n",ans1[i],ans2[i]);
    }
    
}