HDU 1401 solitaire bidirectional BFS

Reprint please indicate the source, thank you http://blog.csdn.net/ACM_cxlove? Viewmode = contents by --- cxlove

The first write bidirectional BFs.

Bidirectional BFS searches from the start and end points. If the intersection points appear, the target point is reachable. This requirement is to arrive within eight steps. Each step has 16 states, and the complexity of 16 ^ 8 is still very high. Because the start and end points are determined, it is very suitable for Bidirectional BFs, that is, you can search for four steps from the start point and the end point respectively. If an intersection occurs, the search is reachable.

Hash is compressed in hexadecimal notation. Each coordinate range is 0-7 and can be expressed in three binary notation. A total of 24 binary notation is used. In hash notation, sort the coordinates. At the beginning, the result of flag [1 <24] is still MLE. Later, I changed it to map.

Although it is written as a BFs, the middle part is written very well.

#include<iostream>#include<cstdio>#include<cstring>#include<queue>#include<cmath>#include<string>#include<vector>#include<algorithm>#include<map>#include<set>#define inf 1<<30#define LL long long#define maxn 1<<24using namespace std;struct Point{int x,y;bool check(){if(x<8&&x>=0&&y<8&&y>=0)return true;return false;}};struct Node{Point pos[4];int step;}s,e;map<int,int>m;map<int,int>::iterator it;int way[4][2]={{0,1},{0,-1},{1,0},{-1,0}};bool cmp(Point n1,Point n2){return n1.x!=n2.x?n1.x<n2.x:n1.y<n2.y;}int get_hash(Point *tmp){int res=0;sort(tmp,tmp+4,cmp);for(int i=0;i<4;i++){res|=(tmp[i].x<<(6*i));res|=(tmp[i].y<<(6*i+3));}return res;}bool bfs(int kind,Node u){Node v;queue<Node>que;u.step=0;if(kind==2){it=m.find(get_hash(u.pos));if(it!=m.end())return true;}m[get_hash(u.pos)]=kind;que.push(u);while(!que.empty()){u=que.front();que.pop();if(u.step>=4)continue;for(int i=0;i<4;i++){for(int j=0;j<4;j++){v=u;v.step++;v.pos[j].x+=way[i][0];v.pos[j].y+=way[i][1];int k;if(!v.pos[j].check())continue;for(k=0;k<4;k++)if(k!=j&&v.pos[k].x==v.pos[j].x&&v.pos[k].y==v.pos[j].y)break;if(k==4){int HASH=get_hash(v.pos);it=m.find(HASH);if(kind==1){if(it==m.end()){m[HASH]=1;que.push(v);}}else{if(it==m.end()){m[HASH]=2;que.push(v);}else if((*it).second==1)return true;}}else{v.pos[j].x+=way[i][0];v.pos[j].y+=way[i][1];if(!v.pos[j].check())continue;for(k=0;k<4;k++)if(k!=j&&v.pos[k].x==v.pos[j].x&&v.pos[k].y==v.pos[j].y)break;if(k==4){int HASH=get_hash(v.pos);it=m.find(HASH);if(kind==1){if(it==m.end()){m[HASH]=1;que.push(v);}}else{if(it==m.end()){m[HASH]=2;que.push(v);}else if((*it).second==1)return true;}}}}}}return false;}int main(){while(scanf("%d%d",&s.pos[0].x,&s.pos[0].y)!=EOF){m.clear();for(int i=1;i<4;i++)scanf("%d%d",&s.pos[i].x,&s.pos[i].y);for(int i=0;i<4;i++)scanf("%d%d",&e.pos[i].x,&e.pos[i].y);for(int i=0;i<4;i++){s.pos[i].x--;s.pos[i].y--;e.pos[i].x--;e.pos[i].y--;}bfs(1,s);printf("%s\n",bfs(2,e)?"YES":"NO");}return 0;}/*2 1 8 8 7 8 1 11 5 8 7 7 7 2 46 5 5 7 5 1 7 17 7 6 5 5 7 6 34 6 5 6 5 5 5 22 2 7 6 5 3 5 24 4 5 5 3 4 7 22 7 7 2 6 5 4 84 2 3 4 5 5 6 62 7 4 4 3 5 4 53 3 5 3 6 2 6 66 3 8 3 6 6 6 83 2 5 3 6 2 6 66 3 8 3 6 6 6 83 4 4 5 5 5 7 53 3 4 3 5 3 6 3*/