1 边界和湖心小岛分别算一个节点。连接全部距离小于D的鳄鱼。时间复杂度O(N2)

2 推断每一个连通图的节点中是否包括边界和湖心小岛,是则Yes否则No

3 冗长混乱的函数參数

#include <stdio.h>
#include <malloc.h>
#include <queue>
#include <math.h> using namespace std; struct Coordinate
{
float x;
float y;
}; bool operator==(Coordinate& a, Coordinate& b)
{
return a.x == b.x && a.y == b.y;
} float DistanceOfPoints(const Coordinate& a, const Coordinate& b)
{
return sqrtf(pow(a.x - b.x, 2) + pow(a.y - b.y, 2));
} void JudgePosition(const int& D, Coordinate* crocodile, const int& i, bool* isCloseToEdge, bool* isCloseToCenter)
{
// 靠近湖岸
if (crocodile[i].x >= 50 - D || crocodile[i].x <= -50 + D ||
crocodile[i].y >= 50 - D || crocodile[i].y <= -50 + D)
{
isCloseToEdge[i] = true;
}
else
{
isCloseToEdge[i] = false;
}
// 靠近湖心小岛
if ( sqrtf(pow(crocodile[i].x, 2) + pow(crocodile[i].y, 2)) <= 7.5 + D)
{
isCloseToCenter[i] = true;
}
else
{
isCloseToCenter[i] = false;
}
} bool IsCloseToEdge(const int& D, const Coordinate& crocodile)
{
return (crocodile.x >= 50 - D || crocodile.x <= -50 + D ||
crocodile.y >= 50 - D || crocodile.y <= -50 + D);
} bool IsCloseToCenter(const int& D, const Coordinate& crocodile)
{
return (sqrtf(pow(crocodile.x, 2) + pow(crocodile.y, 2)) <= 7.5 + D);
} int* CreateMatrixGraph(const int& N)
{
int* graph = (int*) malloc(sizeof(int) * N * N);
for (int i = 0;i < N * N; i++)
{
graph[i] = 0;
}
return graph;
} bool IsMatrixConnected(const int& a, const int& b, int* graph, const int& N)
{
if (a == b)
{
return false;
}
return (graph[a * N + b]);
} void MatrixConnect(const int& a, const int& b, int* graph, const int& N)
{
if (IsMatrixConnected(a, b, graph, N))
{
printf("ERROR : %d AND %d ALREADY CONNECTED\n", a, b);
return;
}
if (a == b)
{
printf("ERROR : THE SAME VERTICE\n");
return;
}
graph[a * N + b] = 1;
graph[b * N + a] = 1;
} void GetAdjoinVertice(const int& vertice, int* graph, int* adjoinVertice, int N)
{
int currentIndex = 0;
for (int i = 0; i < N; i++)
{
if (graph[vertice * N + i] == 1)
{
adjoinVertice[currentIndex++] = i;
}
}
} void DFS(int* graph, const int& vertice, bool* isVisited, int N, bool* result)
{
//printf("%d ", vertice);
isVisited[vertice] = true;
if (vertice == N - 2)
{
result[0] = true;
}
if (vertice == N - 1)
{
result[1] = true;
} int* adjoinVertice = (int*) malloc(sizeof(int) * N);
for (int i = 0; i < N; i++)
{
adjoinVertice[i] = -1;
}
GetAdjoinVertice(vertice, graph, adjoinVertice, N); int i = 0;
while (adjoinVertice[i] != -1)
{
if (!isVisited[adjoinVertice[i]] /*&& DistanceOfPoints(crocodile[vertice], crocodile[i]) <= D*/)
{
DFS(graph, adjoinVertice[i], isVisited, N, result);
}
i++;
}
free(adjoinVertice);
} void BFS(int* graph, int vertice, bool* isVisited, int N)
{
queue<int> t;
t.push(vertice);
isVisited[vertice] = true; while (!t.empty())
{
int currentVertice = t.front();
t.pop();
printf("%d ", currentVertice); int* adjoinVertice = (int*) malloc(sizeof(int) * N);
for (int i = 0; i < N; i++)
{
adjoinVertice[i] = -1;
}
GetAdjoinVertice(currentVertice, graph, adjoinVertice, N);
int i = 0;
while (adjoinVertice[i] != -1)
{
if (!isVisited[adjoinVertice[i]])
{
t.push(adjoinVertice[i]);
isVisited[adjoinVertice[i]] = true;
}
i++;
}
}
} bool MatrixComponentsSearch(int* graph, bool* isVisited, int N, bool* result, int function = 1)
{
for (int i = 0; i < N; i++)
{
if (!isVisited[i])
{
if (function == 1)
{
//printf("{ ");
DFS(graph, i, isVisited, N, result);
if (result[0] == true && result[1] == true)
{
return true;
}
result[0] = false;
result[1] = false;
}
else
{
//printf("{ ");
BFS(graph, i, isVisited, N);
//printf("}\n");
}
}
}
return false;
} int main(void)
{
int N;
int D;
scanf("%d %d", &N, &D);
int nodeCount = N + 2;
Coordinate* crocodile = (Coordinate*) malloc(sizeof(Coordinate) * nodeCount);
bool* isVisited = (bool*) malloc(sizeof(bool) * N); for (int i = 0; i < N; i++)
{
scanf("%f %f", &crocodile[i].x, &crocodile[i].y); }
crocodile[N].x = 0;
crocodile[N].y = 0;
crocodile[N + 1].x = -1;
crocodile[N + 1].y = -1;
// 一共N个鳄鱼。N是湖心小岛。N+1是岸边
int* graph = CreateMatrixGraph(N + 2);
// 连接距离小于D的鳄鱼
for (int i = 0; i < N; i++)
{
if (IsCloseToCenter(D, crocodile[i]))
{
MatrixConnect(i, N, graph, nodeCount);
}
if (IsCloseToEdge(D, crocodile[i]))
{
MatrixConnect(i, N + 1, graph, nodeCount);
}
for (int j = i + 1; j < N; j++)
{
if (DistanceOfPoints(crocodile[i], crocodile[j]) <= D)
{
MatrixConnect(i, j, graph, nodeCount);
}
}
} bool result[2];
result[0] = false;
result[1] = false;
if (MatrixComponentsSearch(graph, isVisited, nodeCount, result))
{
printf("Yes");
}
else
{
printf("No");
} return 0;
}

05-图2. Saving James Bond - Easy Version (25)的更多相关文章

  1. pat05-图2. Saving James Bond - Easy Version (25)

    05-图2. Saving James Bond - Easy Version (25) 时间限制 200 ms 内存限制 65536 kB 代码长度限制 8000 B 判题程序 Standard 作 ...

  2. PTA 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 Bon ...

  3. 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 Bon ...

  4. Saving James Bond - Easy Version (MOOC)

    06-图2 Saving James Bond - Easy Version (25 分) This time let us consider the situation in the movie & ...

  5. Saving James Bond - Easy Version 原创 2017年11月23日 13:07:33

    06-图2 Saving James Bond - Easy Version(25 分) This time let us consider the situation in the movie &q ...

  6. PAT Saving James Bond - Easy Version

    Saving James Bond - Easy Version This time let us consider the situation in the movie "Live and ...

  7. 06-图2 Saving James Bond - Easy Version

    题目来源:http://pta.patest.cn/pta/test/18/exam/4/question/625 This time let us consider the situation in ...

  8. 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 Bon ...

  9. 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 Bon ...

随机推荐

  1. Protocol Buffer Xcode 正确使用思路 成功安装 Xcode7.1

    1. 下载protobuf编译工具 序列化是将数据转换为一个特定的类 http://pan.baidu.com/s/1qWrxHxU 下载解压,它不是用来放在你的项目里 2.打开终端 依次输入并等待指 ...

  2. ASP.NET用SQL Server中的数据来生成JSON字符串

    原文引自:  作者: 缺水的海豚  来源: 博客园  发布时间: 2010-09-21 21:47  阅读: 6136 次  推荐: 0   原文链接   [收藏] 摘要:ExtJs用到的数据内容基本 ...

  3. php 与 ajax 获取123的案例

    同事问我,咱们从数据库里面获取数据,用ajax的方式展示到前台页面.啥都不说了,动手写个案例吧. 1,建立一个页面: <!DOCTYPE html PUBLIC "-//W3C//DT ...

  4. 利用C++ RAII技术自动回收堆内存

    在C++的编程过程中,我们经常需要申请一块动态内存,然后当用完以后将其释放.通常而言,我们的代码是这样的: 1: void func() 2: { 3: //allocate a dynamic me ...

  5. 安装 php

    1.yum安装php yum install php 2.配置 apache 支持 php a.找到httpd.conf find / -name  httpd.conf b.编辑 httpd.con ...

  6. Spring4.0学习笔记(7) —— 通过FactoryBean配置Bean

    1.实现Spring 提供的FactoryBean接口 package com.spring.facoryBean; import org.springframework.beans.factory. ...

  7. (转)QT常用快捷键

    F1        查看帮助F2        跳转到函数定义(和Ctrl+鼠标左键一样的效果)Shift+F2    声明和定义之间切换F4        头文件和源文件之间切换Ctrl+1     ...

  8. .ctor,.cctor 以及 对象的构造过程

    摘要: .ctor,.cctor 以及 对象的构造过程.ctor:简述:构造函数,在类被实例化时,它会被自动调用.当C#的类被编译后,在IL代码中会出现一个名为.ctor的方法,它就是我们的构造函数, ...

  9. 关于$_SERVER 常量 HTTP_X_FORWARDED_HOST与 HTTP_HOST的问题

    今天在看ecshop的源码,发现了用$_SERVER['HTTP_X_FORWARDED_HOST']来判断主机的地址,就目前来说很多人都是直接通过$_SERVER['HTTP_HOST']来判断的, ...

  10. HTML&CSS基础学习笔记1.32-选择器是什么

    选择器是什么 选择器是CSS样式为了定位页面上的任意元素的一种方法. 选择器主要分为:元素标签选择器.通用选择器.类选择器.ID选择器.属性选择器.组合选择器.伪类选择器.伪元素选择器. 先做个了解, ...