···这题,跟着题意用BFS就好了,防止超时,用个map记录一下出现过的字符串就解决了。(写的快,所以代码有点丑)
#include<map>
#include<iostream>
#include<cstdio>
#include<queue>
#include<string>
#include<cstring>
using namespace std;
#define ll long long
#define up(i,a,n) for(int i=a;i<=n;i++)
struct node
{
string s,s1,s2;
int cnt;
};
string a,b,c;
int bfs(node a)
{ map<string,int>mp;
queue<node>q;
q.push(a);
while(!q.empty())
{
node ss=q.front();q.pop();
if(ss.s==c)return ss.cnt;
if(mp[ss.s]!=0)continue;
mp[ss.s]++;
string s2;
int flag=1;
int cnt1=0,cnt2=0;
for(int i=0;i<(ss.s1.size()+ss.s2.size());i++)
{
if(flag)
{
s2+=ss.s2[cnt2++];
flag=false;
}
else
{
s2+=ss.s1[cnt1++];
flag=true;
}
}
string s4,s5;
for(int i=0;i<s2.size()/2;i++)
{
s4+=s2[i];
}
for(int i=s2.size()/2;i<s2.size();i++)
{
s5+=s2[i];
}
node s6;
s6.s=s2;
s6.s1=s4;
s6.s2=s5;
s6.cnt=ss.cnt+1;
q.push(s6);
}
return -1;
}
int main()
{
int t;
scanf("%d",&t);
for(int cas=1;cas<=t;cas++)
{
int n;
scanf("%d",&n);
cin>>a>>b>>c;
node aa;
aa.s1=a;
aa.s2=b;
aa.cnt=0;
aa.s="";
int ans=bfs(aa);
printf("%d %d\n",cas,ans);
}
}