Saving James Bond - Hard Version（30 分）
This time let us consider the situation in the movie "Live and Let Die" in which James Bond, the world‘s most famous spy, was captured by a group of drug dealers. He was sent to a small piece of land at the center of a lake filled with crocodiles. There he performed the most daring action to escape -- he jumped onto the head of the nearest crocodile! Before the animal realized what was happening, James jumped again onto the next big head... Finally he reached the bank before the last crocodile could bite him (actually the stunt man was caught by the big mouth and barely escaped with his extra thick boot).
Assume that the lake is a 100 by 100 square one. Assume that the center of the lake is at (0,0) and the northeast corner at (50,50). The central island is a disk centered at (0,0) with the diameter of 15. A number of crocodiles are in the lake at various positions. Given the coordinates of each crocodile and the distance that James could jump, you must tell him a shortest path to reach one of the banks. The length of a path is the number of jumps that James has to make.
Input Specification:
Each input file contains one test case. Each case starts with a line containing two positive integers N (≤100), the number of crocodiles, and D, the maximum distance that James could jump. Then N lines follow, each containing the (x,y) location of a crocodile. Note that no two crocodiles are staying at the same position.
Output Specification:
For each test case, if James can escape, output in one line the minimum number of jumps he must make. Then starting from the next line, output the position (x,y) of each crocodile on the path, each pair in one line, from the island to the bank. If it is impossible for James to escape that way, simply give him 0 as the number of jumps. If there are many shortest paths, just output the one with the minimum first jump, which is guaranteed to be unique.
Sample Input 1:
17 15
10 -21
10 21
-40 10
30 -50
20 40
35 10
0 -10
-25 22
40 -40
-30 30
-10 22
0 11
25 21
25 10
10 10
10 35
-30 10
Sample Output 1:
4
0 11
10 21
10 35
Sample Input 2:
4 13
-12 12
12 12
-12 -12
12 -12
Sample Output 2:
0
1 #include<stdio.h>
2 #include<math.h>
3 #include<stdlib.h>
4 #define MAX 101
5 #define IsLand 15.0/2;
6 int visited[MAX];
7 int Path[MAX];
8 int N;
9 double R;
10
11 typedef struct Position{
12 double x;
13 double y;
14 int flag;
15 }Position;
16
17 typedef struct Queue* ListQ;
18 typedef struct Queue
19 {
20 int Data[MAX];
21 int rear;
22 int front;
23 }Queue;
24
25 typedef struct Stack* StackQ;
26 typedef struct Stack
27 {
28 int Data[MAX];
29 int Top;
30 }Stack;
31
32 Position G[101];
33 ListQ P=NULL;
34 StackQ Head=NULL;
35
36 ListQ Creat()
37 {
38 ListQ P=(ListQ)malloc(sizeof(Queue));
39 P->rear=P->front=0;
40 return P;
41 }
42
43 void AddQ(ListQ P,int X)
44 {
45 if((P->rear+1)%MAX==P->front)
46 {
47 printf("队列满\n");
48 return;
49 }
50 P->rear=(P->rear+1)%MAX;
51 P->Data[P->rear]=X;
52 return;
53 }
54
55 int IsEmpty(ListQ P)
56 {
57 return P->rear==P->front;
58 }
59
60 int DeleteQ(ListQ P)
61 {
62 if(P->rear==P->front)
63 {
64 printf("队列空\n");
65 return -1;
66 }
67 P->front=(P->front+1)%MAX;
68 return P->Data[P->front];
69 }
70
71 StackQ CreatS()
72 {
73 StackQ P=(StackQ)malloc(sizeof(Stack));
74 P->Top=-1;
75 return P;
76 }
77
78 void Push(StackQ P,int X)
79 {
80 if (P->Top==MAX-1)
81 {
82 printf("满\n");
83 return;
84 }
85 else
86 {
87 P->Data[++P->Top]=X;
88 return;
89 }
90 }
91
92 int IsEmptyS(StackQ P)
93 {
94 return P->Top==-1;
95 }
96
97 int Pop(StackQ P)
98 {
99 if (P->Top==-1)
100 {
101 printf("空\n");
102 return -1;
103 }
104 else
105 return (P->Data[P->Top--]);
106 }
107
108
109 int JumpOut(int v)
110 {
111 int a,b,c;
112 if(G[v].x<0)
113 a=G[v].x+50;
114 else
115 a=50-G[v].x;
116
117 if(G[v].y<0)
118 b=G[v].y+50;
119 else
120 b=50-G[v].y;
121 if(a>b)
122 c=b;
123 else
124 c=a;
125 if(c>R)
126 return 0;
127 else
128 return 1;
129 }
130
131 int FirstJump(int v)
132 {
133 return sqrt((G[v].x*G[v].x)+(G[v].y*G[v].y))<=R+IsLand; //加上小岛本身的距离
134 }
135
136 int MinFirstJump(int v)
137 {
138 return (G[v].x*G[v].x)+(G[v].y*G[v].y); //加上小岛本身的距离
139 }
140
141 int Jump(int i,int v)
142 {
143 return sqrt((G[v].x-G[i].x)*(G[v].x-G[i].x)+(G[v].y-G[i].y)*(G[v].y-G[i].y))
144 <=R;
145 }
146
147 void BFS()
148 {
149 int v;
150 while(!IsEmpty(P))
151 {
152 int i;
153 int v=DeleteQ(P);
154 if(JumpOut(v))
155 {
156 G[v].flag=1;
157 }
158 for(i=0;i<N;i++)
159 {
160 if(visited[i]==-1 && Jump(i,v))
161 {
162 visited[i]=visited[v]+1;
163 Path[i]=v;
164 AddQ(P,i);
165 }
166 }
167 }
168 return;
169 }
170
171 void Sava007()
172 {
173 int v;
174 for(v=0;v<N;v++)
175 {
176 if(visited[v]==-1 && FirstJump(v))
177 {
178 visited[v]=1;
179 Path[v]=N;
180 AddQ(P,v);
181 }
182 }
183 BFS();
184 return;
185 }
186
187 void Out()
188 {
189 int v;
190 int sum=0,k;
191 int min=1000,minSum=0;
192 for (v = 0; v < N; v++)
193 {
194 if(G[v].flag==1)
195 {
196 sum++;
197 k=v;
198 }
199 }
200 if (sum==1)
201 {
202 printf("%d\n",visited[k]+1);
203 for (v=k;v<N;v=Path[v])
204 Push(Head,v);
205 while(!IsEmptyS(Head))
206 {
207 v=Pop(Head);
208 printf("%0.0f %0.0f\n",G[v].x,G[v].y);
209 }
210 }
211 else if (sum>1)
212 {
213 for (v = 0; v < N; v++)
214 {
215 if( G[v].flag==1 && visited[v]<=min)
216 {
217 min=visited[v];
218 k=v;
219 }
220 }
221 for (v = 0; v < N; v++)
222 {
223 if( G[v].flag==1 && visited[v]==min)
224 minSum++;
225 }
226 if (minSum==1)
227 {
228 printf("%d\n",visited[k]+1);
229 for (v=k;v<N;v=Path[v])
230 Push(Head,v);
231 while(!IsEmptyS(Head))
232 {
233 v=Pop(Head);
234 printf("%0.0f %0.0f\n",G[v].x,G[v].y);
235 }
236 }
237 else
238 {
239
240 int i=0,j=0,m=0;
241 int MinAnswer[MAX];
242 int Answer[MAX];
243 for (v=k;v<N;v=Path[v])
244 Push(Head,v);
245 while(!IsEmptyS(Head))
246 MinAnswer[i++]=Pop(Head);
247 for (v = 0; v < N; v++)
248 {
249 m=0;
250 if( G[v].flag==1 && visited[v]==min)
251 {
252 for (j=v;j<N;j=Path[j])
253 Push(Head,j);
254 while(!IsEmptyS(Head))
255 Answer[m++]=Pop(Head);
256 if(MinFirstJump(Answer[0])<MinFirstJump(MinAnswer[0]))
257 {
258 for(j=0;j<i;j++)
259 {
260 MinAnswer[j]=Answer[j];
261 }
262 }
263 }
264 }
265 printf("%d\n",visited[k]+1);
266 for(j=0;j<i;j++)
267 printf("%0.0f %0.0f\n",G[MinAnswer[j]].x,G[MinAnswer[j]].y);
268 }
269 }
270 else
271 printf("0\n");
272 }
273
274 int main()
275 {
276 int i;
277 P=Creat();
278 Head=CreatS();
279 scanf("%d %lf",&N,&R); //注意：跳的距离不一定是整数
280 for(i=0;i<MAX;i++)
281 {
282 visited[i]=-1;
283 Path[i]=-1;
284 }
285 for(i=0;i<N;i++)
286 {
287 scanf("%lf %lf",&G[i].x,&G[i].y); //注意：位置不一定是整数
288 G[i].flag=0;
289
290 }
291 if (R>=43)
292 printf("1\n");
293 else
294 {
295 Sava007();
296 Out();
297 }
298 return 0;
299 }
原文地址：https://www.cnblogs.com/tzfs/p/8807486.html