06-图2 Saving James Bond - Easy Version （25 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 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
代码及注释如下：
/*
我感觉此题用bfs好做一些，所以用了bfs...
起点需要特殊处理下,可以假设让007在原点出发,跳跃的距离多加上小岛的半径即可...
然后其他点与队列中的点比较,看看绝对距离是否小于跳跃距离,小就入队...
终止条件就是看看横或纵坐标加上跳跃距离是否可以到达边界...
没想到居然一遍过了....
*/
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <queue>
#include <math.h>
using namespace std;
const int maxn=105;
int n,d;
struct node
{
int x,y;
};
queue<int>q;
node nn[maxn];
int vis[maxn];
void init()
{
memset (vis,0,sizeof(vis));
}
bool Canreach (int x,int y,double dis)
{
double k=sqrt((nn[x].x-nn[y].x)*(nn[x].x-nn[y].x)+(nn[x].y-nn[y].y)*(nn[x].y-nn[y].y));
if(k>dis)
return false;
else
return true;
}
bool bfs ()
{
vis[0]=1;
for (int i=1;i<=n;i++)
{
if(Canreach(0,i,d+15))
{
q.push(i);
vis[i]=1;
}
}
while (!q.empty())
{
int k=q.front();
if(nn[k].x+d>=50||nn[k].y+d>=50)
return true;
q.pop();
for (int i=1;i<=n;i++)
{
if(!vis[i]&&Canreach(k,i,d))
{
q.push(i);
vis[i]=1;
}
}
}
return false;
}
int main()
{
init();
scanf("%d%d",&n,&d);
nn[0].x=0; nn[0].y=0;
for (int i=1;i<=n;i++)
scanf("%d%d",&nn[i].x,&nn[i].y);
if(bfs())
printf("Yes\n");
else
printf("No\n");
return 0;
}