題目來源：http://pta.patest.cn/pta/test/18/exam/4/question/625

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, 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, 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 /* 能夠跳上的鱷魚相當(dāng)于相鄰的點，第一步比較特殊單獨處理 每次DFS代表一種拯救方案，只要存在一種方案即可獲救 */ #include <cstdio> #include <cmath> #include <cstdlib>const double ISLAND_RADIUS = 15.0 / 2; //孤島半徑 const double SQUARE_SIZE = 100.0; //湖（正方形）的大小 const int N = 110; //鱷魚（點）的最大數(shù) typedef struct Point {double x, y; } Position;Position P[N]; bool Visited[N]; int n; double d;void Save007(); bool DFS(int V); bool FirstJump(int V); bool Jump(int V1, int V2); bool IsSave(int V);int main() {scanf("%d%lf", &n, &d);for (int i = 0; i < n; i++)scanf("%lf%lf", &(P[i].x), &(P[i].y));for (int i = 0; i < n; i++)Visited[i] = false;Save007();return 0; }void Save007() {bool IsSave = false;for (int i = 0; i < n; i++){if (!Visited[i] && FirstJump(i)){IsSave = DFS(i);if (IsSave)break;}}if (IsSave)printf("Yes\n");elseprintf("No\n"); }bool DFS(int V) {Visited[V] = true;bool answer = false;if (IsSave(V))return true;for (int i = 0; i < n; i++){if (!Visited[i] && Jump(V, i))answer = DFS(i);if (answer)break;}return answer; }bool IsSave(int V) {return (abs(P[V].x) >= 50 - d)|| (abs(P[V].y) >= 50 - d); }bool FirstJump(int V) {return sqrt(P[V].x * P[V].x + P[V].y * P[V].y)<= d + ISLAND_RADIUS; }bool Jump(int V1, int V2) {return sqrt((P[V1].x - P[V2].x) * (P[V1].x - P[V2].x) +(P[V1].y - P[V2].y) * (P[V1].y - P[V2].y))<= d; }

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轉(zhuǎn)載于:https://www.cnblogs.com/llhthinker/p/4844990.html

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