Uva-321-brute Force enumeration-implicit graph Search

Test instructions: give you n room, there are many lights of the control switch, I room lamp switch in J room, the room without the lamp can not enter, I room and J room if there is no door, also can not enter from I to J, the beginning of the room is 1, and the lamp is open, asked if you can go to the last room N, And at this time the other room lights are closed. If there are multiple solutions, the sequence of operations with the smallest number of steps is output.

Range: n<=10,

Problem-Solving ideas: Set the status to, the current room number + the current other room lights status. So the total state is n*2^n, the maximum value is 10*1024, and the bare enumeration.

Note that each enumeration of this topic must be encoded from the room 1-n, otherwise WA, and, also, PE WA. Use a bit mark the status of the room light.

#include <iostream>#include<map>#include<memory.h>#include<stdio.h>#include<string>#include<queue>#include<vector>using namespacestd;Const intMAXN = One;classnode{ Public: Vector<string>step; intcur; intCurstepnum; intlights; Node () {//Step.reserve (65535);    }    BOOL operator< (Constnode& node)Const    {        returnCurstepnum >Node.curstepnum; };};intCONN[MAXN][MAXN];intCONTRO[MAXN][MAXN];intR, D, s;intstates[Ten][1025];p riority_queue<Node>Q;intperlights[Ten] = {    1<<0,1<<1,1<<2,1<<3,1<<4,1<<5 ,    1<<6,1<<7,1<<8,1<<9 };voidRead () {intSS, EE;  for(inti =0; I < d;i++) {cin>> SS >>ee; //start with 0CONN[SS-1][ee-1] =1; Conn[ee-1][SS-1] =1; }     for(inti =0; I < s;i++) {cin>> SS >>ee; Contro[ss-1][ee-1] =1;    }}node BFS () {node node; Node.cur=0; Node.curstepnum=0; Node.lights=1;    Q.push (node); //init node, only 0 rooms are lit.states[0][1] =1;  while(Q.empty () = =false) {node=Q.top ();        Q.pop (); intCurindex =node.cur; //cur Lights State        intCurlights =node.lights; intCurstepnum =Node.curstepnum; //Judging is not the end        if(Curindex = = R-1) && Curlights = = Perlights[r-1])        {            returnnode; }            //enumerating the status of Lights         for(inti =0; I <Ten; i++)        {            if(Contro[curindex][i] = =0)                Continue; if(i = =Curindex)Continue; //control of the lamp            intNextcontrolightindex =i; stringdesc =""; intNextlight =0; if((Curlights & perlights[nextcontrolightindex]) = =Perlights[nextcontrolightindex]) {                //turn off the lights in next roomNextlight = (curlights ^Perlights[nextcontrolightindex]); if(States[curindex][nextlight] = =1)                    //Repeat                    Continue; Desc="-Switch off light in room"+ std::to_string (Nextcontrolightindex +1); Desc+="."; }            Else            {                //turn on the lights in next roomNextlight = Curlights |Perlights[nextcontrolightindex]; if(States[curindex][nextlight] = =1)                    //Repeat                    Continue; Desc="-Switch on light in room"+ std::to_string (Nextcontrolightindex +1); Desc+=".";            } Node NewNode; Newnode.curstepnum= Curstepnum +1; Newnode.cur=Curindex; Newnode.lights=Nextlight; Newnode.step=Node.step;            NewNode.step.push_back (DESC); //push to queueQ.push (NewNode); States[curindex][nextlight]=1; }        //end Curindex enum Lights State//Start Door         for(inti =0; I <Ten; i++)        {            if(Curindex = =i)Continue; if(Conn[curindex][i] = =0)                Continue; intNextdoor =i; if((Perlights[nextdoor] & curlights) = =0)                //the lights are off and cannot be entered.                Continue; if(States[nextdoor][curlights] = =1)                //I've seen it, can't go in                Continue; //the light is on, it can go inNode NewNode; Newnode.cur=Nextdoor; Newnode.curstepnum= Curstepnum +1; Newnode.lights=curlights; Newnode.step=Node.step; stringdesc ="-Move to room"+ std::to_string (Nextdoor +1); Desc+=".";            NewNode.step.push_back (DESC); States[nextdoor][curlights]=1;        Q.push (NewNode); }} node.cur= -1; returnnode;}intMain () {intt =0;  while(CIN >> R >> D >>s) {if(r = = d && d = = S && r = =0)             Break; MEMSET (Conn,0,sizeof(conn)); memset (Contro,0,sizeof(conn)); memset (states,0,sizeof(states));  while(Q.empty () = =false) Q.pop ();                Read (); cout<<"Villa #"; cout<< T +1<<Endl; if(r = =1) {cout<<"The problem can solved in"<<0<<"steps:"<<Endl; }        Else{node node=BFS (); //cout << "test" << Endl;            if(Node.cur = =-1) {cout<<"The problem cannot be solved."<<Endl; }            Else{cout<<"The problem can solved in"<< Node.curstepnum <<"steps:"<<Endl; Vector<string> b =Node.step;  for(inti =0; I <= b.size ()-1; i++) cout<< B[i] <<Endl;; }} t++; cout<<Endl; }}

Uva-321-brute Force enumeration-implicit graph Search