"Uvalive 7364" Robots (Reverse thinking + search)

"Uvalive 7364" Robots (Reverse thinking + search)

Main topic:
n a robot on a coordinate 0~n-1.
There are two buttons that allow the robot at each location to go to the point specified by the button (∈[0,n−1] \in [0,n-1]).

Ask if you can go through a few operations to get all the robots to a point.

Considering that the final state is all robots at one point, consider the 22 robot interactions, starting with the state (i,i) to reverse the search, the state of the search (A, b) can be achieved (I,I) after several operations of a or B..

So traverse all the robot pairs, if 22 can reach the same point, eventually there must be a way all to a point.

So just bfs a little bit on the line. Starting with the DFS results tle ...

The code is as follows:

#include <iostream> #include <cmath> #include <vector> #include <cstdlib> #include <cstdio > #include <climits> #include <ctime> #include <cstring> #include <queue> #include <stack&
Gt  #include <list> #include <algorithm> #include <map> #include <set> #define LL Long Long #define Pr pair<int,int> #define FREAD (CH) freopen (CH, "R", stdin) #define FWRITE (CH) freopen (CH, "w", stdout) using namespace s
td
const int INF = 0X3F3F3F3F;
const int mod = 1E9+7;
Const double EPS = 1e-8;

const int MAXN = 1123;
Vector <int> VC[2][MAXN];

BOOL VIS[MAXN][MAXN];
    void BFs (int a,int b) {queue <Pr> q;

    Q.push (Pr (b));
    Pr p;

    VIS[A][B] = Vis[b][a] = 1;
        while (!q.empty ()) {p = Q.front ();

        Q.pop ();
        A = P.first;

        b = P.second; for (int i = 0, i < 2; ++i) for (int j = 0, J < vc[i][a].size (); ++j) for (int k = 0; K &lt ;Vc[i][b].size (); ++K) if (!vis[vc[i][a][j]][vc[i][b][k]]) {VIS[VC[I][A][J]]
                        [Vc[i][b][k]] = vis[vc[i][b][k]][vc[i][a][j]] = 1;
                    Q.push (Pr (Vc[i][b][k],vc[i][a][j]));
    }}} bool solve (int n) {int x;

    for (int j = 0, J < 2; ++j) for (int i = 0; i < n; ++i) vc[j][i].clear ();
        for (int j = 0; j < 2; ++j) for (int i = 0; i < n; ++i) {scanf ("%d", &x);
    Vc[j][x].push_back (i);
    } memset (Vis,0,sizeof (VIS));

    for (int i = 0; i < n; ++i) BFS (i,i); for (int i = 0, i < n; ++i) for (int j = 0; J < N; ++j) if (!vis[i][j]) return FAL

    Se
return true;
    } int main () {//fread ("");

    Fwrite ("");

    int t,k,n;

    scanf ("%d", &t);
        while (t--) {scanf ("%d%d", &k,&n); printf ("%d%s\n", K,solve (n)?
    "YES": "NO"); } return 0; }