06-图2 Saving James Bond

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 ($\le 100$), the number of crocodiles, and D, the maximum distance that James could jump. ThenN 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 2025 -15-25 288 4929 15-35 -25 2827 -29-8 -28-20 -35-25 -20-13 29-30 15-35 4012 12

Sample Output 1:

Yes

Sample Input 2:

4 13-12 1212 12-12 -1212 -12

Sample Output 2:

No

解析：用邻接矩阵或邻接表会比较麻烦，必须先计算所有结点间的距离再和007的跳跃距离比较，才能构建图的边。使用Jump函数和FirstJump函数来判断两个结点之间是否有边，可以省去存储边的麻烦。

#include <stdio.h>#include <stdlib.h>#include <math.h>int N, D;int Visited[105];struct Crocodile {int x, y;} cro[105];double ComputeDistance( int a, int b ) {double x = abs( cro[a].x - cro[b].x ); double y = abs( cro[a].y - cro[b].y );return sqrt( x * x + y * y );}int FirstJump( struct Crocodile dot ) {double temp = dot.x * dot.x + dot.y * dot.y;double dis = sqrt( temp );double cap = 7.5 + D;//第一次跳要加上小岛的半径if( cap >= dis ) return 1;else return 0;}int Jump( int a, int b ) {double dis = ComputeDistance(a, b);if( D >= dis) return 1;else return 0;}//v点是否能直接跳上岸边int IsSafe( int v ) {int x = abs( cro[v].x );int y = abs( cro[v].y );if( x + D >= 50 || y + D >= 50 )return 1;else return 0;}//DFS改编int DFS(int v) {int answer = 0;Visited[v] = 1;if( IsSafe(v) ) answer = 1;else {for(int i = 0; i < N; i++)if( !Visited[i] && Jump(v, i) ) {   //Jump来判断是否是邻接的结点（与邻接矩阵有着相同的作用）answer = DFS(i);if(answer == 1) break;}}return answer;}//计算连通集个数 改编void Save007() {int answer = 0;for(int i = 0; i < N; i++) {if (!Visited[i] && FirstJump(cro[i])) {answer = DFS(i);if (answer == 1) break;}}if (answer == 1) printf("Yes");else printf("No");}int main(){scanf("%d %d", &N, &D);for(int i = 0; i < N; i++)scanf("%d %d", &cro[i].x, &cro[i].y);Save007();system("pause");return 0;}