07-图5 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 (≤), the number of crocodiles, and D, the maximum distance that James could jump. Then N lines follow, each containing the ( 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 ( 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 <cstdio>
  2 #include <cmath>
  3 #include <algorithm>
  4 #include <queue>
  5 #include <cstring>
  6 #include <stack>
  7 using namespace std;
  8 
  9 const int diameter = 15;
 10 const int maxn = 102;
 11 int D, Ncro;
 12 
 13 struct coordinates {
 14     int x, y;
 15 } cro_pos[maxn];
 16 
 17 
 18 struct JNode {
 19     int final;
 20     int length, start;
 21     
 22 } ans_set[maxn];
 23 
 24 
 25 coordinates Origin;
 26 
 27 void init(int Ncro);
 28 bool jumpJudge(coordinates a, coordinates b);
 29 bool arrive(int i);
 30 bool firstJudge(coordinates a, coordinates b);
 31 
 32 bool reachBank(int vertex);
 33 double dist(coordinates a, coordinates b);
 34 
 35 
 36 
 37 
 38 void weighted(int, int);
 39 bool cmp(JNode a, JNode b);
 40 bool findShortest(int );
 41 
 42 int Distance[maxn];
 43 int Parent[maxn];
 44 int NearBank[maxn];
 45 
 46 
 47 
 48 int main() {
 49     scanf("%d %d", &Ncro, &D);
 50     init(Ncro);
 51     stack<coordinates> st;
 52     Distance[0] = 0;
 53     
 54     if(reachBank(0) ) {NearBank[0] = true;}
 55     
 56     for(int V=1; V<=Ncro; V++) {
 57         
 58         if ( firstJudge(cro_pos[V], Origin) ) {
 59             Distance[V] = 1;
 60             Parent[V] = 0;
 61             weighted(V, Ncro);
 62         }
 63     }
 64     
 65     if( !findShortest(Ncro) ) printf("0\n");
 66     else {
 67         printf("%d\n", ans_set[0].length + 1);
 68         
 69         for(int i=ans_set[0].final; i; i = Parent[i] ) {
 70             st.push( cro_pos[i] );
 71         }
 72         while(!st.empty()) {
 73             coordinates tmp = st.top();
 74             st.pop();
 75             
 76             printf("%d %d\n", tmp.x, tmp.y);
 77             
 78             
 79         }
 80         
 81         
 82     }
 83     
 84 }
 85 
 86 bool findShortest(int Ncro) {
 87     int num = 0, j;
 88     bool flag = false;
 89     
 90     for(int i=0; i<=Ncro; i++) {
 91         if(NearBank[i]) {
 92             flag = true;
 93             ans_set[num].final = i;
 94             ans_set[num].length = Distance[i];
 95             
 96             for(j=i; Parent[j] > 0; j = Parent[j]);
 97             ans_set[num].start = j;
 98             
 99             num++;
100         }
101     }
102     
103     sort(ans_set, ans_set+num, cmp);
104     
105     
106     return flag;
107     
108 }
109 
110 bool cmp(JNode a, JNode b) {
111     if (a.length != b.length) {
112         return a.length < b.length;
113     }
114     else return ( dist(cro_pos[a.start], Origin) < dist(cro_pos[b.start], Origin) );
115 }
116 
117 
118 void weighted(int S, int Ncro) {
119     //reset struct ans[s]
120     queue<int> Q;
121     int W;
122     
123     Q.push(S);
124     
125     while(!Q.empty()) {
126         W = Q.front();
127         Q.pop();
128         
129         if(reachBank(W) ) {NearBank[W] = true; continue;}
130         
131         for(int i=1; i<=Ncro; i++ ) {
132             
133             if( jumpJudge( cro_pos[W], cro_pos[i] ) )
134                 if( Distance[i] == -1 ) {
135                     Distance[i] = Distance[W] + 1;
136                     Parent[i] = W;
137                     Q.push(i);
138                 }
139                 
140             
141         }
142     }
143     
144     
145 }
146 
147 bool reachBank(int vertex) {
148     int ans = 0;
149     
150     if (fabs(cro_pos[vertex].x - 50) <= D) {
151         ans =  1;
152     }
153     
154     if (fabs(cro_pos[vertex].x + 50) <= D) {
155         ans =  1;
156     }
157     
158     if (fabs(cro_pos[vertex].y + 50) <= D) {
159         ans =  1;
160     }
161     
162     if (fabs(cro_pos[vertex].y - 50) <= D) {
163         ans = 1;
164     }
165     
166     return ans;
167 }
168 
169 
170 
171 
172 void init(int Ncro) {
173     for (int i=1; i<=Ncro; i++) {
174         scanf("%d%d", &cro_pos[i].x, &cro_pos[i].y);
175     }
176     
177     Origin.x = 0;
178     Origin.y = 0;
179     memset( Distance, -1, sizeof(Distance) );
180     memset( Parent, -1, sizeof(Parent) );
181 }
182 
183 bool jumpJudge(coordinates a, coordinates b) {
184     if ( dist(a, b) <= D) {
185         return true;
186     }
187     else return false;
188 }
189 
190 
191 bool firstJudge(coordinates a, coordinates ori) {
192     
193     
194     if ( dist(a, ori) <= D + diameter/2 and dist(a, ori) > diameter/2) {
195         
196         return  true;
197     }
198     return false;
199 }
200 
201 double dist(coordinates a, coordinates b) {
202     return sqrt( pow(a.x - b.x, 2) + pow(a.y - b.y, 2) ) ;
203 }

 

posted @ 2019-05-28 10:30  Acoccus  阅读(239)  评论(0编辑  收藏  举报