06-图2 Saving James Bond - Easy Version-优快云博客

本文链接：https://blog.youkuaiyun.com/qq_18262861/article/details/108279407

在电影《生死关头》中，詹姆斯邦德被毒贩困于鳄鱼湖中央小岛，通过跳跃鳄鱼头部逃生。本博客解析了如何用算法判断邦德能否成功逃脱，涉及坐标、距离计算及路径寻找。

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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:

Sample Output 1:

Yes

Sample Input 2:

Sample Output 2:

No

代码实现

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#define MaxSize 100
typedef int Position;
typedef int Vertex;
struct VNode
{
    Position X;
    Position Y;
}Node[MaxSize];
int Visited[MaxSize]={0};
/*判断是否可以上岸*/
int Is_Safe(Vertex V,int D){
    int Flag=0;
    if (((abs(Node[V].X)+D)>=50)||((abs(Node[V].Y)+D)>=50))
        Flag=1;
    return Flag;
}
/*判断是否可以跳到下一个顶点*/
int Jump(Vertex V,Vertex W,int D){
    int Flag=0;
    /*两点间的距离是否小于D*/
    if (sqrt(((Node[V].X-Node[W].X)*(Node[V].X-Node[W].X))+((Node[V].Y-Node[W].Y)*(Node[V].Y-Node[W].Y)))<=D)
        Flag=1;
    return Flag;
}
/*深度遍历结点*/
int DFS(Vertex V,int N,int D){
    Position W;
    int Answer=0;
    Visited[V]=1;
    /*判断该结点是否可以上岸*/
    if (Is_Safe(V,D))   Answer=1;
    else{
        for (W=0; W<N; W++){
            if (!Visited[W]&&Jump(V,W,D))
            {
                Answer=DFS(W,N,D);
                if (Answer==1)  break; //上岸，停止循环
            }
        }
    }
    return Answer;
}
/*判断是否可以由中心跳到该结点*/
int First_Jump(Vertex V,int D){
    int Flag=0;
    /*（0,0）为圆心，（D+中心半径）为半径的圆*/
    if (((Node[V].X)*(Node[V].X)+(Node[V].Y*Node[V].Y))<=((7.5+D)*(7.5+D)))
        Flag=1;
    return Flag;
}
void Svae007(int N,int D){
    int Answer=0;
    Vertex V;
    for (V=0; V<N; V++){
        if ((Visited[V]!=1)&&(First_Jump(V,D)==1))
        {
            Answer=DFS(V,N,D);
            if (Answer==1)  break;
        }  
    }
    if (Answer==1)  printf("Yes");
    else printf("No");

}
int main()
{
    int N,D;
    scanf("%d %d",&N,&D);
    for (int i = 0; i < N; i++)
        scanf("%d %d",&Node[i].X,&Node[i].Y);
    Svae007(N,D);
    return 0;
}