HDU 4444 离散化+bfs 注意细节

时间：2015-05-18 16:11:24 阅读：141 评论：0 收藏：0 [点我收藏+]

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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?
技术分享

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

0 5 5 8

0 0 0 10

0 5 5 8

-2 1 0 5

0 0 0 0

Sample Output

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 }

HDU 4444 离散化+bfs 注意细节

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原文地址：http://www.cnblogs.com/qq1012662902/p/4511997.html

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