題目鏈接：hdu 4444 Walk

題意：

有n個矩形，現在biaoge要從一個點走到另外一個點上，不能穿過矩形，但可以沿著邊走。

問最少需要多少次轉彎才能到達。

題解：

由於數據很大，需要離散化一下，然後就是一個bfs，用三維的vis來記錄最好的狀態。

細節比較多，需要考慮延邊走時的拐角方向。

  1 #include<bits/stdc++.h>
  2 #define mst(a,b) memset(a,b,sizeof(a))
  3 #define F(i,a,b) for(int i=(a);i<=(b);++i)
  4 using namespace std;
 
  5 typedef pair<int,int>P;
  6 
  7 const int N=507,inf=1e9,all=1e8+7;
  8 int mp[N][N],st[2],en[2],vis[4][N][N];
  9 int hshx[N],hshy[N],hx_ed,hy_ed;
 10 int n,NN,MM,dir[][2]={1,0,0,1,-1,0,0,-1};
 11 vector<int>X,Y;
 12 struct SQR
 13 {
 14     P point[2];
 15     void in(int a,int b,int c,int
 
 d)
 16     {
 17         point[0].first=min(a,c);
 18         point[0].second=min(b,d);
 19         point[1].first=max(a,c);
 20         point[1].second=max(b,d);
 21     }
 22     void get()
 23     {
 24         point[0].first=lower_bound(X.begin(),X.end(),point[0].first)-X.begin()+1;
 25         point[1
 
].first=lower_bound(X.begin(),X.end(),point[1].first)-X.begin()+1;
 26         point[0].second=lower_bound(Y.begin(),Y.end(),point[0].second)-Y.begin()+1;
 27         point[1].second=lower_bound(Y.begin(),Y.end(),point[1].second)-Y.begin()+1;
 28     }
 29 }sqr[100];
 30 
 31 struct Node
 32 {
 33     int x,y,turn,d;
 34     Node(int a=0,int b=0,int c=0,int d_=0):x(a),y(b),turn(c),d(d_){}
 35 };
 36 
 37 int check(int x,int y,int prex,int prey,int i,int prei)
 38 {
 39     if(x<1||x>NN||y<1||y>MM)return 0;
 40     if(mp[x][y]==1)
 41     {
 42         if((mp[x][y+1]==2&&mp[x][y-1]==2)||
 43           (mp[x+1][y]==2&&mp[x-1][y]==2))return 0;
 44         if((mp[x-1][y-1]==2&&mp[x+1][y+1]==2))
 45         {
 46             if(prei==0&&i!=1)return 0;
 47             if(prei==3&&i!=2)return 0;
 48             if(prei==1&&i!=0)return 0;
 49             if(prei==2&&i!=3)return 0;
 50         }
 51         if((mp[x-1][y+1]==2&&mp[x+1][y-1]==2))
 52         {
 53             if(prei==0&&i!=3)return 0;
 54             if(prei==3&&i!=0)return 0;
 55             if(prei==1&&i!=2)return 0;
 56             if(prei==2&&i!=1)return 0;
 57         }
 58     }
 59 
 60     return 1;
 61 }
 62 
 63 int bfs()
 64 {
 65     queue<Node>Q;
 66     F(k,0,3)F(i,0,NN+1)F(j,0,MM+1)vis[k][i][j]=inf;
 67     F(k,0,3)vis[k][st[0]][st[1]]=0;
 68     Q.push(Node(st[0],st[1],0,-1));
 69     while(!Q.empty())
 70     {
 71         Node now=Q.front();Q.pop();
 72         if(now.d!=-1&&now.turn>vis[now.d][now.x][now.y])continue;
 73         if(now.x==en[0]&&now.y==en[1])continue;
 74         F(i,0,3)
 75         {
 76             int tx=now.x+dir[i][0],ty=now.y+dir[i][1];
 77             int tturn=now.turn;
 78             if(now.d!=-1&&now.d!=i)tturn++;
 79             if(!check(now.x,now.y,tx,ty,i,now.d)||vis[i][tx][ty]<=tturn||mp[tx][ty]>1)continue;
 80             Q.push(Node(tx,ty,tturn,i));
 81             vis[i][tx][ty]=tturn;
 82         }
 83     }
 84     int ans=inf;
 85     F(i,0,3)ans=min(ans,vis[i][en[0]][en[1]]);
 86     return ans;
 87 }
 88 
 89 int main()
 90 {
 91     while(scanf("%d%d%d%d",st,st+1,en,en+1),st[0]||st[1]||en[1]||en[0])
 92     {
 93         st[0]+=all,st[1]+=all,en[0]+=all,en[1]+=all;
 94         st[0]*=2,st[1]*=2,en[0]*=2,en[1]*=2;
 95         hshx[hx_ed=1]=st[0];hshx[++hx_ed]=en[0];
 96         hshy[hy_ed=1]=st[1];hshy[++hy_ed]=en[1];
 97         scanf("%d",&n);
 98         X.clear(),Y.clear();
 99         F(i,1,n)
100         {
101             int a,b,c,d;
102             scanf("%d%d%d%d",&a,&b,&c,&d);
103             a+=all,b+=all,c+=all,d+=all;
104             a*=2,b*=2,c*=2,d*=2;
105             sqr[i].in(a,b,c,d);
106             hshx[++hx_ed]=a,hshx[++hx_ed]=c;
107             hshy[++hy_ed]=b,hshy[++hy_ed]=d;
108         }
109         sort(hshx+1,hshx+1+hx_ed),hx_ed=unique(hshx+1,hshx+1+hx_ed)-hshx-1;
110         sort(hshy+1,hshy+1+hy_ed),hy_ed=unique(hshy+1,hshy+1+hy_ed)-hshy-1;
111         X.push_back(-inf),X.push_back(hshx[1]);
112         F(i,2,hx_ed)
113         {
114             if(hshx[i]>hshx[i-1]+1)X.push_back(hshx[i-1]+1);
115             X.push_back(hshx[i]);
116         }
117         X.push_back(inf);
118         Y.push_back(-inf),Y.push_back(hshy[1]);
119         F(i,2,hy_ed)
120         {
121             if(hshy[i]>hshy[i-1]+1)Y.push_back(hshy[i-1]+1);
122             Y.push_back(hshy[i]);
123         }
124         Y.push_back(inf);
125         F(i,1,n)sqr[i].get();
126         st[0]=lower_bound(X.begin(),X.end(),st[0])-X.begin()+1;
127         st[1]=lower_bound(Y.begin(),Y.end(),st[1])-Y.begin()+1;
128         en[0]=lower_bound(X.begin(),X.end(),en[0])-X.begin()+1;
129         en[1]=lower_bound(Y.begin(),Y.end(),en[1])-Y.begin()+1;
130         NN=X.size(),MM=Y.size();
131         F(i,0,NN+1)F(j,0,MM+1)mp[i][j]=0;
132         F(i,1,n)
133         {
134             int x1=sqr[i].point[0].first,x2=sqr[i].point[1].first;
135             int y1=sqr[i].point[0].second,y2=sqr[i].point[1].second;
136             F(j,x1+1,x2-1)F(k,y1+1,y2-1)mp[j][k]=2;
137             F(j,x1,x2)F(k,y1,y2)if(mp[j][k]==0)mp[j][k]=1;
138         }
139         int ans=bfs();
140         printf("%d\n",ans==inf?-1:ans);
141     }
142     return 0;
143 }

hdu 4444 Walk(離散化+BFS)