06-图2 Saving James Bond - Easy Version

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. 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

#include<stdlib.h>

typedef float Elemetype;
typedef struct v {
	int n;
	struct v* next;
}V,*qV;

typedef struct point {
	Elemetype x;
	Elemetype y;
	int states;
}Point,*qPoint;

typedef struct stack {
	int *n;	//*qV的编号
	int top;
}Stack,*qStack;

//构造栈
void CreateStack(qStack *q,int numbers) {
	*q = (Stack *)malloc(sizeof(Stack));
	(*q)->n = (int *)malloc(sizeof(int)*numbers);
	(*q)->top = -1;
}

//出栈
int Pop(qStack *q) {
	int N;
	if ((*q)->top != -1) {
		N = (*q)->n[(*q)->top];
		(*q)->top--;
	}
	return N;
}
//入栈
void Push(qStack *q, int N) {
	(*q)->top++;
	(*q)->n[(*q)->top] = N;
}

//构造矩阵表
void Create(qV **q,int numbers) {
	*q = (qV *)malloc(sizeof(qV)*numbers);
	for (int i = 0; i < numbers; i++)
	{
		V *a = (V *)malloc(sizeof(V));
		a->n = i;
		//a->states = 0;
		a->next = NULL;
		(*q)[i] = a;
	}
}

//判断点集是否能连通
int isLink(Point q, Point p,float d) {
	if (p.x==0&&p.y==0) {
		float D = (q.x - p.x)*(q.x - p.x) + (q.y - p.y)*(q.y - p.y);
		if (D <= (d+15/2.0) * (d + 15/2.0)) {
			return 1;
		}
		else {
			return 0;
		}
	}
	else {
		float D = (q.x - p.x)*(q.x - p.x) + (q.y - p.y)*(q.y - p.y);
		if (D <= d * d) {
			return 1;
		}
		else {
			return 0;
		}
	}
}

//判断能否逃脱
int isPao(Point q,float D) {
	if (q.x >= (50 - D) || q.x <= (D - 50) || q.y >= (50-D) || q.y <= (D - 50)) {
		return 1;
	}
	else {
		return 0;
	}
}
//向矩阵表插入节点
void Insert(qV **q,int i,int n) {
	if (i != n) {
		qV s = (qV)malloc(sizeof(V));
		s->n = i;
		s->next = (*q)[n]->next;
		(*q)[n]->next = s;
	}
}

//构造点集
qPoint Creat(qPoint q,int number) {
	q = (Point *)malloc(sizeof(Point)*number);
	return q;
}

//开始遍历图
int Throgh(qStack *stack,qV *q,int numbers,qPoint p,float D) {
	int N;
	qV s;
	s = q[0];	//起点的周围点都放入栈中
	while (s->next != NULL) {
		s = s->next;
		Push(stack, s->n);
		p[(s->n) - 1].states = 1;
	}
	//栈不为空时
	while ((*stack)->top!=-1)
	{
		N = Pop(stack);
		if (isPao(p[N-1], D) == 1) {
			return 1;
		}
		s = q[N];
		while (s->next != NULL) {	//队列都入栈
			s = s->next;
			if (p[(s->n) - 1].states != 1) {
				Push(stack, (s)->n);
				p[(s->n) - 1].states = 1;
				
			}
		}
	}
	return 0;
}


int main()
{
	//读入的数以及距离
	int numbers,D;
	scanf("%d %d", &numbers, &D);
	
	//构造矩阵表
	qV *a = NULL;
	Create(&a, numbers+1);

	//创建栈
	qStack stack=NULL;
	CreateStack(&stack,numbers);

	//原点
	Point p0;
	p0.x = 0;
	p0.y = 0;
	p0.states = 0;

	//构造点集
	qPoint q = NULL;
	q=Creat(q,numbers);
	int X, Y;
	for (int  i = 0; i < numbers; i++)
	{
		scanf("%d %d", &X, &Y);
		q[i].x = X;
		q[i].y = Y;
		q[i].states = 0;

		//起始点能走的路径都插入
		if (isLink(q[i],p0 , D)==1) {
			Insert(&a, i+1, 0);
		}

	}
	

	//从第一个点开始插入能连通的路径
	for (int i = 1; i <= numbers; i++)
	{
		for (int j = 0; j < numbers; j++)
		{
			if (isLink(q[j], q[i-1], D) == 1) {
				Insert(&a, j + 1, i);
			}
		}
	}

	int flag= Throgh(&stack,  a,  numbers, q,  D);
	if (flag == 1) {
		printf("Yes");
	}
	else {
		printf("No");
	}
    return 0;
}