uva321 bfs搜索

这道题系统没法提交，所以不知道能否AC，但我感觉程序没啥问题，比较简单，可以跑几个测试用例比对一下。

题意容易理解，思路 是将所在的房间号和r个房间的开灯情况作为一个状态，这样最多会有10*2^10种状态，并不多。然后就是bfs宽度遍历了，要注意一点是人在某个房间中不可以把该房间的灯关上，所以我在输入的灯开关数组时候就做了一个判断，如果两个数字相同，也即控制该房间灯的开关就在这个房间时，开关是无效的（很不符合常理~~）这样就保证了之后开关灯的合法性。

#include<iostream>#include<cstring>#include<cstdio>using namespace std;const int MAXSIZE = 60000;const int N=11;struct state{int room[N];int location,behave;};int r,d,s,front,rear,step,prestate[MAXSIZE];bool flag,door[N][N],swit[N][N],vis[N][1025];state target,S[MAXSIZE];void init(){memset(swit,0,sizeof(swit));memset(door,0,sizeof(door));memset(S,0,sizeof(S));memset(vis,0,sizeof(vis));S[0].room[1]=1;S[0].location=1;vis[1][1]=1;target.location=r;memset(target.room,0,sizeof(target.room));target.room[r]=1;}int caldex(state s){int dex=0,exp=1;for (int i=1;i<=r;i++){dex=dex+s.room[i]*exp;exp=exp << 1 ;}return dex;}bool compare(state s){if (s.location==target.location){for (int i=1;i<=r;i++){if (s.room[i]!=target.room[i])return 0;}return 1;}return 0;}void bfs(){front=0;rear=1,step=0,flag=0;while (front<rear){step++;int tmprear=rear;while (front<tmprear){state cs = S[front];int i;for (i=1;i<=r;i++){if (swit[cs.location][i]){state &ns = S[rear];ns=S[front];ns.room[i]=!ns.room[i];if (compare(ns)){flag=1;prestate[rear]=front;ns.behave=i*10+(ns.room[i]==1?1:2);//1:turn on;2:turn offreturn;}int dex=caldex(ns);if (!vis[ns.location][dex]){vis[ns.location][dex]=1;prestate[rear]=front;ns.behave=i*10+(ns.room[i]==1?1:2);//1:turn on;2:turn offrear++;}}}for (i=1;i<=r;i++){if (door[cs.location][i]&&cs.room[i]){state &ns=S[rear];ns=cs;ns.location=i;int dex=caldex(ns);if (!vis[ns.location][dex]){vis[ns.location][dex]=1;prestate[rear]=front;ns.behave=3*100+i;rear++;}}}front++;}}}void print(state s){cout<<"- ";if (s.behave>300){cout<<"Move to room "<<s.behave%300<<"."<<endl;}else{if (s.behave%10==1){cout<<"Switch on light in room "<<s.behave/10<<"."<<endl;}else{cout<<"Switch off light in room "<<s.behave/10<<"."<<endl;}}}int main(){int col=0;while (cin>>r>>d>>s&&r){col++;init();for (int i=0;i<d;i++){int m,n;cin>>m>>n;door[m][n]=door[n][m]=1;}for (int i=0;i<s;i++){int m,n;cin>>m>>n;if (m!=n)swit[m][n]=1;}bfs();cout<<"Villa #"<<col<<endl;if (flag){cout<<"The problem can be solved in "<<step<<" steps:"<<endl;int *result = new int[step];int pre=rear;for (int i=0;i<step;i++){result[i]=pre;pre=prestate[pre];}for (int j=step-1;j>=0;j--){print(S[result[j]]);}delete result;}else cout<<"The problem cannot be solved."<<endl;cout<<endl;getchar();}return 0;}