双向BFS统计

Hdu1195

两个四位密码 问你最少多少步能到达

/*Huyyt*/
#include<bits/stdc++.h>
#define mem(a,b) memset(a,b,sizeof(a))
#define pb push_back
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int mod = 1e9 + ;
const int gakki =  +  +  +  + 1e9;
const int MAXN = 2e5 + , MAXM = 2e5 + ;
/*int to[MAXM << 1], nxt[MAXM << 1], Head[MAXN], ed = 1;
inline void addedge(int u, int v)
{
        to[++ed] = v;
        nxt[ed] = Head[u];
        Head[u] = ed;
}*/
struct node
{
        int now[];
        int step;
} startpoint, endpoint, cnt, to;
void print(node x)
{
        for (int i = ; i <= ; i++)
        {
                cout << x.now[i];
        }
}
int finalans;
bool flag = false;
int vis[][][][];
char s[], e[];
queue<node> que[];
int stepsum[];
node get_change(node x, int aim)
{
        node cur = x;
        swap(cur.now[aim], cur.now[aim + ]);
        return cur;
}
node get_next(node x, int aim, int y)
{
        x.now[aim] += y;
        if (x.now[aim] > )
        {
                x.now[aim] = ;
        }
        else if (x.now[aim] < )
        {
                x.now[aim] = ;
        }
        return x;
}
void get_vis(node x, int y)
{
        vis[x.now[]][x.now[]][x.now[]][x.now[]] = y;
}
void check(node x, int y)
{
        if (vis[x.now[]][x.now[]][x.now[]][x.now[]] == (y ^ ))
        {
                finalans = min(finalans, stepsum[] + stepsum[] + );
                flag = true;
        }
        else if (vis[x.now[]][x.now[]][x.now[]][x.now[]] == -)
        {
                que[y].push(x);
                get_vis(x, y);
        }
}
void doit(int x)
{
        while (que[x].front().step == stepsum[x] && que[x].size())
        {
                cnt = que[x].front();
                que[x].pop();
                for (int i = ; i <= ; i++)
                {
                        if (i != )
                        {
                                to = get_change(cnt, i);
                                to.step = cnt.step + ;
                                check(to, x);
                        }
                        to = get_next(cnt, i, );
                        to.step = cnt.step + ;
                        //cout << x << " ", print(cnt), cout << " ", print(to), cout << endl;
                        check(to, x);
                        to = get_next(cnt, i, -);
                        to.step = cnt.step + ;
                        //cout << x << " ", print(cnt), cout << " ", print(to), cout << endl;
                        check(to, x);
                }
        }
        stepsum[x]++;
}
int main()
{
        int T;
        scanf("%d", &T);
        while (T--)
        {
                flag = false;
                scanf("%s", s + ), scanf("%s", e + );
                stepsum[] = stepsum[] = ;
                for (int i = ; i <= ; i++)
                {
                        for (int j = ; j <= ; j++)
                        {
                                for (int k = ; k <= ; k++)
                                {
                                        for (int w = ; w <= ; w++)
                                        {
                                                vis[i][j][k][w] = -;
                                        }
                                }
                        }
                }
                finalans = INT_MAX;
                for (int i = ; i <= ; i++)
                {
                        while (que[i].size())
                        {
                                que[i].pop();
                        }
                }
                startpoint.step = , endpoint.step = ;
                for (int i = ; i <= ; i++)
                {
                        startpoint.now[i] = s[i] - '';
                        endpoint.now[i] = e[i] - '';
                }
                //print(startpoint), print(endpoint);
                que[].push(startpoint), que[].push(endpoint);
                get_vis(startpoint, ), get_vis(endpoint, );
                while (!flag)
                {
                        if (que[].size() > que[].size())
                        {
                                doit();
                        }
                        else
                        {
                                doit();
                        }
                }
                printf("%d\n", finalans);
        }
        return ;
}

//Hdu1195

Hdu1401

8X8的棋盘上有四个棋子 问你能不能在八步之内把一个状态转移到另一一个状态