題目←

qbxt裡雙向BFS的例題……
網上好多題解用的是STL黑科技？

抱著練碼力的心態手寫了佇列+find+replace;
順便map大法好啊

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cctype>
#include<cstring>
#include<map>
using namespace std;
const int MAXN = 20000 + 60;
struct zt
{
    string a,b;
}ch[MAXN];
string
 
 a,b,s,t;
int cnt;
string q1[MAXN],q2[MAXN];
int head1,head2,tail1,tail2;
bool same(string x,string y)
{
    if(x.length() != y.length())return false;
    int len = x.length();
    for(int i = 0;i < len;i ++)
    {
        if(x[i] != y[i])return false;
    }
    return true;
}
int check(string a,string
 
 b,int st)
{
    if(b.length() > a.length())return -1;
    int len1 = a.length(),len2 = b.length();
    for(int i = st;i < len1;i ++)
    {
        if(a[i] == b[0])
        {
            bool flag = true;
            for(int j = 0;j < len2;j ++)
            {
                if(a[i + j] != b[j])
                {
                    flag = false
 
;
                    break;
                }
            }
            if(flag)return i;
        }
    }
    return -1;
}
map <string,bool> used1,used2;
map <string,int> step1,step2;
string re(string a,string b,string c,int pos)
{
    string ans;
    ans = "";
    int pos2 = pos + b.length();
    int loc = 0;
    while(loc < pos)
    {
        ans += a[loc];
        loc ++;
    }
    int len = c.length();
    for(int i = 0;i < len;i ++)
    {
        ans += c[i];
        loc ++;
    }
    len = a.length();
    for(int i = pos2;i < len;i ++)
    {
        ans += a[i];
        loc ++;
    }
    return ans;
}
int T,flag;
bool F;
int main()
{
    //cin >> T;
    cin >> s >> t;
    while(cin >> a >> b)
    {
        ch[++ cnt].a = a;
        ch[cnt].b = b;
    }
    q1[tail1 ++] = s;q2[tail2 ++] = t;
    used1[s] = true;
    used2[t] = true;
    while(!same(q1[tail1 - 1],q2[tail2 - 1]))
    {

        F = 0;
        for(int i = 1;i <= cnt;i ++)
        {
            for(int j = 0;j < q1[head1].length();j ++)
            {
                int loc = check(q1[head1],ch[i].a,j);
                if(loc != -1)
                {
                    j = loc + 1;
                    string tmp = re(q1[head1],ch[i].a,ch[i].b,loc);
                    if(used1[tmp])continue;
                    F = 1;
                    used1[tmp] = true;
                    step1[tmp] = step1[q1[head1]] + 1;
                    q1[tail1 ++] = tmp;
                    if(used2[tmp])
                    {
                        printf("%d",step1[tmp] + step2[tmp]);
                        return 0;
                    }
                    //cout << flag << tmp << '\n';
                }
            }

        }
        flag ^= 1;
        for(int i = 1;i <= cnt;i ++)
        {
            for(int j = 0;j < q2[head2].length();j ++)
            {
                int loc = check(q2[head2],ch[i].b,j);
                if(loc != -1)
                {
                    j = loc + 1;
                    string tmp = re(q2[head2],ch[i].b,ch[i].a,loc);
                    if(used2[tmp])continue;
                    F = 1;
                    used2[tmp] = true;
                    step2[tmp] = step2[q2[head2]] + 1;
                    q2[tail2 ++] = tmp;
                    if(used1[tmp])
                    {
                        printf("%d",step1[tmp] + step2[tmp]);
                        return 0;
                    }
                    //cout << flag << tmp << '\n';
                }
            }
        }
        flag ^= 1;
        head1 ++;
        head2 ++;
        if(head1 == tail1 || head2 == tail2)
        {
            printf("NO ANSWER!\n");
            return 0;
        }
    }

    return 0;
}

其實……這是假的雙向BFS
感謝wwq大佬指出的bug

按程式碼的搜尋順序，先擴充套件點2；再擴充套件點1
此時搜到相同解，路徑長度為4
事實上點1的兄弟節點還可能擴展出其它更優解，如圖

路徑長度為3
綜上，正確寫法為先將某佇列中同一深度的點擴充套件完，再擴充套件另一佇列，保證深度嚴格遞增。