UVA-12569 Planning mobile Robot on the Tree (Easy Version) (bfs+ State compression)

The main idea: an undirected graph, a robot, a number of stones, each move can only move to the connected node, and only one node can have the same and one thing (robot or stone), to find a robot from the specified position to another specified position of the minimum number of steps, the output move step.

Analysis of the problem: the position of the robot and the location of the stone to mark the status of the state, the number of States has a maximum of 15*2^15. Wide search of the.

The code is as follows:

# include<iostream># include<cstdio># include<string># include<queue># include<vector ># include<cstring># include<algorithm>using namespace std;struct edge{int to,nxt;};    struct node{int t,u,sta;    string path;        Node (int _t,int _u,int _s,string _p): t (_t), U (_u), STA (_s), Path (_p) {} BOOL operator < (const node &a) Const {    Return t>a.t; }};    Edge e[32];int n,cnt,head[16],vis[15][1<<15];void Add (int u,int v) {e[cnt].to=v;    E[cnt].nxt=head[u]; head[u]=cnt++;}    void BFs (int s,int st,int ed) {priority_queue<node>q;    memset (vis,0,sizeof (VIS));    Vis[s][st]=1;    Q.push (Node (0,s,st, ""));        while (!q.empty ()) {node u=q.top ();        Q.pop ();            if (u.u==ed) {printf ("%d\n", u.t);            for (int i=0;i<u.path.size (); i+=2) printf ("%d%d\n", u.path[i]-' a ' +1,u.path[i+1]-' a ' + 1);        return; } for (int i=0;i<n;++i) {if (u.sta& (1&Lt;<i)) {for (int j=head[i];j!=-1;j=e[j].nxt) {int v=e[j].to;                    if (u.sta& (1<<v)) continue;                    if (V==U.U) continue;                    int ns=u.sta^ (1<<i);                    Ns|= (1&LT;&LT;V);                        if (!vis[u.u][ns]) {vis[u.u][ns]=1;                        String P=u.path;                        p+= (char) (i+ ' a '), p+= (char) (v+ ' a ');                    Q.push (Node (u.t+1,u.u,ns,p));            }}}} for (int i=head[u.u];i!=-1;i=e[i].nxt) {int v=e[i].to;            if (u.sta& (1<<v)) continue;                if (!vis[v][u.sta]) {vis[v][u.sta]=1;                String P=u.path;                p+= (char) (u.u+ ' a '), p+= (char) (v+ ' a ');            Q.push (Node (u.t+1,v,u.sta,p)); }}} printf (" -1\n");} int MaiN () {int t,a,b,s,t,st,m,cas=0;    scanf ("%d", &t);        while (t--) {st=cnt=0;        memset (head,-1,sizeof (head));        scanf ("%d%d%d%d", &n,&m,&s,&t);        --s,--t;            while (m--) {scanf ("%d", &a);        St|= (1<< (A-1));            } for (int i=1;i<n;++i) {scanf ("%d%d", &a,&b);            Add (a-1,b-1);        Add (B-1,a-1);        } printf ("Case%d:", ++cas);        BFS (s,st,t);    if (T) printf ("\ n"); } return 0;}

　　

UVA-12569 Planning mobile Robot on the Tree (Easy Version) (bfs+ State compression)