(c++详解)07-图5 Saving James Bond - Hard Version(详细解释)Input Specification:Output Specification:分析

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我是靠谱客的博主陶醉星月，最近开发中收集的这篇文章主要介绍(c++详解)07-图5 Saving James Bond - Hard Version(详细解释)Input Specification:Output Specification:分析，觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

本博文源于浙江大学《数据结构》，题目中拯救007是图（上)课后习题的翻版，需要大家给出007具体的跳跃路径，如果让路径具有记忆性，容器并不可少，所以我才用了c++，背离了c语言，郭伟老师说stl是神兵利器，如果以后同学们刷算题还是学会使用神兵利器较好呀！
先把题目放出来！

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 a shortest path to reach one of the banks. The length of a path is the number of jumps that James has to make.

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, if James can escape, output in one line the minimum number of jumps he must make. Then starting from the next line, output the position (x,y) of each crocodile on the path, each pair in one line, from the island to the bank. If it is impossible for James to escape that way, simply give him 0 as the number of jumps. If there are many shortest paths, just output the one with the minimum first jump, which is guaranteed to be unique.

Sample Input 1:

Sample Output 1:

分析

如果直接上来做题目还是比较难整的，需要大家认真思考一下，老师一开始做最短路径的搜索匹配时，就说了这是可以解决007指路问题，所以这方面是关键，本代码采用广度优先搜索鳄鱼能跳，配合堆栈+队列容器，函数要判断当前踩的鳄鱼是否立马上岸，并且跟新最短路径，还要记录踩过的鳄鱼！第一步跳跃也要选择好，如果一开始就能上岸，那就立马输出，这些都是可以考虑被设计的，完整代码如下：

#include "iostream"
#include "math.h"
#include "queue"
#include "stack"
#include "algorithm"
using namespace std;
int n, m;
#define MINLEN 42.5
struct Pointer {
int x;
int y;
}point[101];
bool answer = false; /* 记录007能否安全逃生~~ */
bool visited[101] = { false }; /* 判断当前点是否被访问过 */
int path[101] = {-1}; /* 记录跳跃过程中踩过的鳄鱼 */
bool isSave(int x) { /* 判断从当前点能否跳到岸上 */
if ((point[x].x - m <= -50) || (point[x].x + m >= 50) || (point[x].y - m <= -50) || (point[x].y + m >= 50))
return true;
return false;
}
bool jump(int x, int y) { /* 判断2个点距离是否在跳跃能力内 */
int p1 = pow(point[x].x - point[y].x, 2);
int p2 = pow(point[x].y - point[y].y, 2);
int r = m * m;
if (p1 + p2 <= r)
return true;
return false;
}
int firstJump(int x) {
/* 当007处于孤岛时 第一次可以选择跳的鳄鱼 因为第一次判断能否跳跃的计算方法与后面dfs不相同 所以要单独写 */
int p1 = pow(point[x].x, 2);
int p2 = pow(point[x].y, 2);
int r = (m + 7.5) * (m + 7.5);
if (p1 + p2 <= r) {
return p1+p2;
}
return 0;
}
bool cmp(int a,int b) {
return firstJump(a) < firstJump(b);
}
void bfs() { /* 用bfs来判断最少要踩几个小鳄鱼才能上岸 */
int b[101];
queue<int>q;
/* 将第一步能踩到的鳄鱼按距离从小到大的顺序进队列~ 因为输出结果要保证在踩的鳄鱼数量相等的情况下 输出第一步距离最短的~~*/
for (int i = 0; i < n; i++) {
b[i] = i;
}
sort(b, b + n, cmp); /* 按照第一步的距离排序~~~ */
int last;
for (int i = 0; i < n; i++) {
if (firstJump(b[i])) { /* 能跳上去！ */
q.push(b[i]);
visited[b[i]] = true; /* 指向当前层数最后一个数~ */
last = b[i];
}
}
int step = 2;
/* 记录最少要跳跃的次数 */
int tail;
while (!q.empty()) {
int p = q.front();
q.pop();
if (isSave(p)) {
int k = 1;
stack<int> s;
cout << step << endl;
while (k < step) {
//cout << point[p].x << " " << point[p].y << endl;
s.push(p);
p = path[p];
k++;
}
while (!s.empty()) {
p = s.top();
s.pop();
cout << point[p].x << " " << point[p].y << endl;
}
return;
}
for (int i = 0; i < n; i++) {
if (!visited[i] && jump(p, i)) { /* 没踩过并且能跳到 */
q.push(i);
path[i] = p; /* 记得当前进队节点的父节点~ */
visited[i] = true;
tail = i; /* 指向下一层的最后一个元素 */
}
}
if (last == p) { /* 即将进入下一层~ */
step += 1;
last = tail;
}
}
if (q.empty()) { /* 如果队列为空
说明没跳出去啊~ */
cout << "0" << endl;
}
}
int main() {
cin >> n >> m;
for (int i = 0; i < n; i++) {
cin >> point[i].x >> point[i].y;
}
if (m >= MINLEN) { /* 可以直接从孤岛上提到岸上 直接输出 */
cout << "1" << endl;
return 0;
}
bfs();
return 0;
}