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 + 7;
const int gakki = 5 + 2 + 1 + 19880611 + 1e9;
const int MAXN = 2e5 + 5, MAXM = 2e5 + 5;
/*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[5];
        int step;
} startpoint, endpoint, cnt, to;
void print(node x)
{
        for (int i = 1; i <= 4; i++)
        {
                cout << x.now[i];
        }
}
int finalans;
bool flag = false;
int vis[10][10][10][10];
char s[10], e[10];
queue<node> que[2];
int stepsum[2];
node get_change(node x, int aim)
{
        node cur = x;
        swap(cur.now[aim], cur.now[aim + 1]);
        return cur;
}
node get_next(node x, int aim, int y)
{
        x.now[aim] += y;
        if (x.now[aim] > 9)
        {
                x.now[aim] = 1;
        }
        else if (x.now[aim] < 1)
        {
                x.now[aim] = 9;
        }
        return x;
}
void get_vis(node x, int y)
{
        vis[x.now[1]][x.now[2]][x.now[3]][x.now[4]] = y;
}
void check(node x, int y)
{
        if (vis[x.now[1]][x.now[2]][x.now[3]][x.now[4]] == (y ^ 1))
        {
                finalans = min(finalans, stepsum[0] + stepsum[1] + 1);
                flag = true;
        }
        else if (vis[x.now[1]][x.now[2]][x.now[3]][x.now[4]] == -1)
        {
                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 = 1; i <= 4; i++)
                {
                        if (i != 4)
                        {
                                to = get_change(cnt, i);
                                to.step = cnt.step + 1;
                                check(to, x);
                        }
                        to = get_next(cnt, i, 1);
                        to.step = cnt.step + 1;
                        //cout << x << " ", print(cnt), cout << " ", print(to), cout << endl;
                        check(to, x);
                        to = get_next(cnt, i, -1);
                        to.step = cnt.step + 1;
                        //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 + 1), scanf("%s", e + 1);
                stepsum[0] = stepsum[1] = 0;
                for (int i = 1; i <= 9; i++)
                {
                        for (int j = 1; j <= 9; j++)
                        {
                                for (int k = 1; k <= 9; k++)
                                {
                                        for (int w = 1; w <= 9; w++)
                                        {
                                                vis[i][j][k][w] = -1;
                                        }
                                }
                        }
                }
                finalans = INT_MAX;
                for (int i = 0; i <= 1; i++)
                {
                        while (que[i].size())
                        {
                                que[i].pop();
                        }
                }
                startpoint.step = 0, endpoint.step = 0;
                for (int i = 1; i <= 4; i++)
                {
                        startpoint.now[i] = s[i] - '0';
                        endpoint.now[i] = e[i] - '0';
                }
                //print(startpoint), print(endpoint);
                que[0].push(startpoint), que[1].push(endpoint);
                get_vis(startpoint, 0), get_vis(endpoint, 1);
                while (!flag)
                {
                        if (que[0].size() > que[1].size())
                        {
                                doit(1);
                        }
                        else
                        {
                                doit(0);
                        }
                }
                printf("%d
", finalans);
        }
        return 0;
}

 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