双向BFS统计

Posted 2020-12-30 aragaki

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;
int finalans;
int dis[5][5][5][5];
char s[10], e[10];
queue<node> que[2];
int stepsum[2];
void get_dis(node x, int y)
{
        dis[x.now[1]][x.now[2]][x.now[3]][x.now[4]] = y;
}
void doit(int x)
{
        while (que[x].front().step == stepsum[x])
        {
                cnt = que[x].front();
                que[x].pop();

        }
        stepsum[x]++;
}
int main()
{
        int T;
        scanf("%d", &T);
        while (T--)
        {
                scanf("%s", s + 1), scanf("%s", e + 1);
                stepsum[0] = stepsum[1] = 0;
                memset(visit, -1, sizeof(visit));
                finalans = INT_MAX;
                qs.clear(), qe.clear();
                startpoint.step = 0, endpoint.step = 1e9;
                for (int i = 1; i <= 4; i++)
                {
                        startpoint.now[i] = s[i] - ‘0‘;
                        endpoint.now[i] = e[i] - ‘0‘;
                }
                que[0].push(startpoint), que[1].push(endpoint);
                if (que[0].size() > que[1].size())
                {
                        doit(1);
                }
                else
                {
                        doit(0);
                }
        }
        return 0;
}

//TO DO

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