hdu 4444 Walk(离散化+BFS)

题目链接：hdu 4444 Walk

题意：

有n个矩形，现在biaoge要从一个点走到另外一个点上，不能穿过矩形，但可以沿着边走。

问最少需要多少次转弯才能到达。

题解：

由于数据很大，需要离散化一下，然后就是一个bfs，用三维的vis来记录最好的状态。

细节比较多，需要考虑延边走时的拐角方向。

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

,st[2],en[2],vis[4]

;
9 int hshx
,hshy
,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 }