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这题比较简单，用了数组。在007跳跃范围内，找到鳄鱼，依次直到岸边 return yes。否则No
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 whether or not he can escape.
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, andD, the maximum distance that James could jump. Then NN 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, print in a line "Yes" if James can escape, or "No" if not.
Sample Input 1:
14 20
25 -15
-25 28
8 49
29 15
-35 -2
5 28
27 -29
-8 -28
-20 -35
-25 -20
-13 29
-30 15
-35 40
12 12
Sample Output 1:
Yes
Sample Input 2:
4 13
-12 12
12 12
-12 -12
12 -12
Sample Output 2:
No
1 #include <iostream> 2 #include <cstdio> 3 using namespace std; 4 #define MaxSize 100 5 6 struct crocodiles 7 { 8 float x; 9 float y; 10 }; 11 struct crocodiles cro[MaxSize]; 12 13 int D; //007可以跳的距离 14 int N; //鳄鱼数量 15 bool Visited[MaxSize] = {false}; 16 17 //距离小于 007 可跳距离 返回true 18 bool Jump(int x1, int y1, int x2, int y2) 19 { 20 if( (x2-x1)*(x2-x1) + (y2-y1)*(y2-y1) <= D*D)//两点距离小于可跳距离 21 return true; 22 return false; 23 } 24 25 //如果007跳到岸边 则返回true 26 bool isSafe(int v) 27 { 28 if( (cro[v].x + D >= 50) || (cro[v].y + D >= 50) || (cro[v].y - D <= -50) || (cro[v].x - D <= -50)) 29 return true; 30 else 31 return false; 32 } 33 34 bool DFS ( int V ) 35 { 36 bool answer = false; 37 Visited[V] = true; 38 if( isSafe(V) ) 39 answer = true; 40 else { 41 for(int i = 0; i < N; i++) { 42 if( !Visited[i] && Jump(cro[V].x,cro[V].y,cro[i].x,cro[i].y)) { 43 answer = DFS(i); 44 if(answer == true) 45 break; 46 } 47 } 48 } 49 return answer; 50 } 51 52 //第一跳是否可以跳到 是return true 53 bool firstJump(int x, int y) 54 { 55 if( x*x + y*y <= (7.5+D)*(7.5+D) ) 56 return true; 57 else 58 return false; 59 } 60 61 void save007() 62 { 63 bool answer = false; 64 for(int i = 0; i < N; i++) { 65 if( !Visited[i] && firstJump(cro[i].x,cro[i].y) ) { 66 answer = DFS(i); 67 if( answer == true) 68 break; 69 } 70 } 71 if(answer == true) 72 printf("Yes "); 73 else 74 printf("No "); 75 } 76 77 78 int main() 79 { 80 scanf("%d %d",&N,&D); 81 // if(N < (35/D - 1)) { //N至少为 (50-15) /D -1 82 // printf("No "); 83 // return 0; 84 // } 85 for(int i = 0; i < N; i++) 86 scanf("%f %f",&cro[i].x,&cro[i].y); 87 save007(); 88 return 0; 89 }