Walk
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1972 Accepted Submission(s): 308
Problem Description
Biaoge is planning to walk to amusement park. The city he lives can be abstracted as a 2D plane. Biaoge is at (x1, y1) and the amusement park is at (x2, y2). There are also some rectangle buildings. Biaoge can only walk parallel to the coordinate axis. Of course Biaoge can’t walk across the buildings.
What’s the minimum number of turns Biaoge need to make?
As the figure above shows, there are 4 buildings and Biaoge need to make at least 3 turns to reach the amusement park(Before walking he can chose a direction freely). It is guaranteed that all the buildings are parallel to the coordination axis. Buildings may contact but overlapping is impossible. The amusement park and Biaoge’s initial positions will not contact or inside any building.
Input
There are multiple test case.
Each test case contains several lines.
The first line contains 4 integers x1, y1, x2, y2 indicating the coordinate of Biaoge and amusement park.
The second line contains one integer N(0≤N≤50), indicating the number of buildings.
Then N lines follows, each contains 4 integer x1, y1, x2, y2, indicating the coordinates of two opposite vertices of the building.
Input ends with 0 0 0 0, you should not process it.
All numbers in the input range from -108 to 108.
Output
For each test case, output the number of least turns in a single line. If Biaoge can’t reach the amusement park, output -1 instead.
Sample Input
0 0 0 10
1
0 5 5 8
0 0 0 10
2
0 5 5 8
-2 1 0 5
0 0 0 0
Sample Output
0
2
Hint
In the first case, Biaoge can walk along the side of building, and no turn needed.
In the second case, two buildings block the direct way and Biaoge need to make 2 turns at least.
题目意思：
有n个矩形，然后两个不在矩形上面的点，求一个点到另一个点路径上最少的拐弯数目（点不能穿过矩形，但是可以在边界上走）。
思路：
离散化所有点，建图，bfs就行了。
可以走
不可以走
代码：
1 #include <cstdio>
2 #include <cstring>
3 #include <algorithm>
4 #include <iostream>
5 #include <vector>
6 #include <queue>
7 #include <cmath>
8 #include <set>
9 #include <bitset>
10 #include <map>
11 using namespace std;
12
13 #define N 400
14 #define inf 999999999
15
16 int max(int x,int y){return x>y?x:y;}
17 int min(int x,int y){return x<y?x:y;}
18 int abs(int x,int y){return x<0?-x:x;}
19
20 int xx[]={-1,0,1,0};
21 int yy[]={0,1,0,-1};
22 int lx, ly;
23 int n;
24
25 struct node{
26 int x1, y1, x2, y2;
27 }a[55];
28 int llx[N], lly[N];
29 map<int,int>mapx, mapy;
30 int stx, sty, endx, endy;
31 int sx, sy, ex, ey;
32 int g[N][N];
33 int dp[N][N][4];
34 bool visited[N][N][4];
35
36 void creat(){ //建图
37 memset(g,0,sizeof(g));
38 int i, j, k;
39 for(i=0;i<n;i++){
40 for(j=mapx[a[i].x1];j<=mapx[a[i].x2];j++){
41 g[j][mapy[a[i].y1]]=2;//printf("11111\n");
42 g[j][mapy[a[i].y2]]=2;
43 }
44 for(j=mapy[a[i].y1]+1;j<mapy[a[i].y2];j++){
45 g[mapx[a[i].x1]][j]=2;
46 g[mapx[a[i].x2]][j]=2;
47 }
48 for(j=mapx[a[i].x1]+1;j<mapx[a[i].x2];j++){
49 for(k=mapy[a[i].y1]+1;k<mapy[a[i].y2];k++){
50 g[j][k]=2;
51 }
52 }
53 }
54 for(i=0;i<=lx;i++){
55 for(j=0;j<=ly;j++){
56 if(j<ly&&!g[i][j]&&g[i][j+1]) g[i][j+1]=1;
57 if(j>0&&!g[i][j]&&g[i][j-1]) g[i][j-1]=1;
58 if(i<lx&&!g[i][j]&&g[i+1][j]) g[i+1][j]=1;
59 if(i>0&&!g[i][j]&&g[i-1][j]) g[i-1][j]=1;
60 }
61 }
62 for(i=1;i<lx;i++){
63 for(j=1;j<ly;j++){
64 if(g[i][j]==2){
65 if(!g[i+1][j+1]&&g[i-1][j]==1&&g[i][j-1]==1) g[i][j]=1;
66 if(!g[i-1][j-1]&&g[i-1][j]==1&&g[i][j-1]==1) g[i][j]=1;
67 if(!g[i+1][j-1]&&g[i-1][j]==1&&g[i][j+1]==1) g[i][j]=1;
68 if(!g[i-1][j+1]&&g[i-1][j]==1&&g[i][j+1]==1) g[i][j]=1;
69 if(!g[i-1][j+1]&&g[i+1][j]==1&&g[i][j-1]==1) g[i][j]=1;
70 if(!g[i+1][j-1]&&g[i+1][j]==1&&g[i][j-1]==1) g[i][j]=1;
71 if(!g[i-1][j-1]&&g[i+1][j]==1&&g[i][j+1]==1) g[i][j]=1;
72 if(!g[i+1][j+1]&&g[i+1][j]==1&&g[i][j+1]==1) g[i][j]=1;
73 }
74 if(g[i][j]==1&&g[i-1][j-1]==2&&g[i+1][j+1]==2&&!g[i+1][j-1]&&!g[i-1][j+1]) g[i][j]=2;
75 if(g[i][j]==1&&!g[i-1][j-1]&&!g[i+1][j+1]&&g[i+1][j-1]==2&&g[i-1][j+1]==2) g[i][j]=2;
76 }
77 }
78 }
79
80 struct G{
81 int x, y, z, f;
82 };
83
84 void bfs(){
85 int i, j, k;
86 queue<G>Q;
87 G p, q;
88 for(i=0;i<=lx;i++){
89 for(j=0;j<=ly;j++){
90 for(k=0;k<4;k++){
91 dp[i][j][k]=inf;
92 }
93 }
94 }
95 memset(visited,false,sizeof(visited));
96 p.x=sx;p.y=sy;p.f=0;
97 for(i=0;i<4;i++){
98 p.z=i;
99 dp[p.x][p.y][p.z]=0;
100 visited[p.x][p.y][p.z]=true;
101 Q.push(p);
102 }
103 while(!Q.empty()){
104 p=Q.front();Q.pop();
105 visited[p.x][p.y][p.z]=false;
106 for(i=0;i<4;i++){
107 q.x=p.x+xx[i];
108 q.y=p.y+yy[i];
109 q.z=i;
110 if(p.f>0&&q.z!=p.f-1) continue; //之前的点是拐角的话，必须按照走特定的方向
111 if(q.x<0||q.x>lx||q.y<0||q.y>ly) continue;
112 if(p.f>0) q.f=0;
113 if(g[q.x][q.y]==2){ //拐角判断
114 if(!g[q.x-1][q.y-1]&&!g[q.x+1][q.y+1]){
115 if(q.z==0) q.f=1+1;
116 else if(q.z==3) q.f=2+1;
117 else if(q.z==1) q.f=0+1;
118 else if(q.z==2) q.f=3+1;
119 }
120 else if(!g[q.x+1][q.y-1]&&!g[q.x-1][q.y+1]){
121 if(q.z==0) q.f=3+1;
122 else if(q.z==3) q.f=0+1;
123 else if(q.z==1) q.f=2+1;
124 else if(q.z==2) q.f=1+1;
125 }
126 else continue;
127 }
128 else q.f=0;
129 if(q.z==p.z){
130 if(dp[q.x][q.y][q.z]>dp[p.x][p.y][p.z]){
131 dp[q.x][q.y][q.z]=dp[p.x][p.y][p.z];
132 if(!visited[q.x][q.y][q.z]){
133 Q.push(q);
134 visited[q.x][q.y][q.z]=true;
135 }
136 }
137 }
138 else{
139 if(dp[q.x][q.y][q.z]>dp[p.x][p.y][p.z]+1){
140 dp[q.x][q.y][q.z]=dp[p.x][p.y][p.z]+1;
141 if(!visited[q.x][q.y][q.z]){
142 Q.push(q);
143 visited[q.x][q.y][q.z]=true;
144 }
145 }
146 }
147 }
148 }
149 }
150
151 main()
152 {
153 int i, j, k;
154 int len1, len2;
155 while(scanf("%d %d %d %d",&stx,&sty,&endx,&endy)==4){
156 if(!stx&&!sty&&!endx&&!endy) break;
157 len1=len2=0;
158 llx[len1++]=stx;
159 llx[len1++]=endx;
160 lly[len2++]=sty;
161 lly[len2++]=endy;
162 scanf("%d",&n);
163 for(i=0;i<n;i++){
164 scanf("%d %d %d %d",&a[i].x1,&a[i].y1,&a[i].x2,&a[i].y2);
165 llx[len1++]=a[i].x1;
166 llx[len1++]=a[i].x2;
167 lly[len2++]=a[i].y1;
168 lly[len2++]=a[i].y2;
169 }
170 sort(llx,llx+len1);
171 sort(lly,lly+len2);
172 lx=ly=1;
173 mapx.clear();
174 mapy.clear();
175 for(i=0;i<len1;i++){
176 if(mapx.find(llx[i])==mapx.end()){
177 mapx[llx[i]]=lx;lx+=2;
178 }
179 }
180 for(i=0;i<len2;i++){
181 if(mapy.find(lly[i])==mapy.end()){
182 mapy[lly[i]]=ly;ly+=2;
183 }
184 }
185 creat();
186 /* for(i=0;i<len1;i++) printf("%d ",mapx[llx[i]]);
187 cout<<endl;
188 for(i=0;i<=lx;i++){
189 for(j=0;j<=ly;j++){
190 printf("%d ",g[i][j]);
191 }
192 cout<<endl;
193 }*/
194 sx=mapx[stx];ex=mapx[endx];
195 sy=mapy[sty];ey=mapy[endy];
196 bfs();
197 int ans=inf;
198 for(i=0;i<4;i++){
199 ans=min(ans,dp[ex][ey][i]);
200 }
201 if(ans!=inf) printf("%d\n",ans);
202 else printf("-1\n");
203 }
204 }
O5YBIE)DJ(TD~SJ`5E309C.png)