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

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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:

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
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:

1
2
3
4
5
4 13
-12 12
12 12
-12 -12
12 -12

Sample Output 2:
No

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#include<iostream>
#include<vector>
#include<map>
#include<utility>
#include<cmath>

using namespace std;

class Graph
{
public:
    Graph(int num, int maxDis):vertexNum(num), maxDistance(maxDis)
    {
        int x = 0, y = 0;
        for(int i = 1; i < num + 1; ++i)
        {
            cin >> x >> y;
            pair<int, int >pos(x, y);
            position.insert(make_pair(i, pos));
        }
        position.insert(make_pair(num + 1, make_pair(50, 50))); //(50, 50)是随便给的，代表岸边，操作中不起作用
        visited.resize(num + 2, 0);
        m_graph.resize(num + 2);
        powMaxDistance = pow(maxDistance, 2);

        CreateGraph();
    };

    void CreateGraph();
    bool BFS(int index);
    bool Escape();  //fail or succeed to escape

private:
    vector<int> visited;
    map<int , pair<int, int> > position;
    int maxDistance;
    int powMaxDistance;     //pow(maxDistance, 2)
    int vertexNum;			// the number of vertex
    vector<vector<int> > m_graph;        //store graph with 1D vector
};

void Graph::CreateGraph()
{
    int StartMaxDistance = pow(7.5 + (float)maxDistance, 2);
    for(int i = 1; i < vertexNum + 1; ++i)  //对起点（james站的直径15那个圆盘单独做处理
    {
        if(pow(position[0].first - position[i].first, 2) +
           pow(position[0].second - position[i].second, 2) <= StartMaxDistance)
         {
            m_graph[0].push_back(i);
            m_graph[i].push_back(0);
         }
    }
    for(int j = 1; j < vertexNum + 1; ++j)  //把鳄鱼位置当作结点，做处理
    {
        for(int i = j + 1; i < vertexNum + 1; ++i)
        {
            if(pow(position[j].first - position[i].first, 2) +
                pow(position[j].second - position[i].second, 2) <= powMaxDistance)
            {
                m_graph[j].push_back(i);
                m_graph[i].push_back(j);  //加上这句不会有重复元素，因为加了if(!visited1[i])
            }
        }
    }
    for(int i = 1; i < vertexNum + 1; ++i)      //判断有无离岸边在规定距离的点，对岸边这个结点做处理
    {
        if(abs(position[i].first  - 50)  <= maxDistance
        || abs(position[i].second - 50)  <= maxDistance
        || abs(position[i].first  + 50)  <= maxDistance
        || abs(position[i].second + 50)  <= maxDistance)
        {
            m_graph[i].push_back(vertexNum + 1);
            m_graph[vertexNum + 1].push_back(i);
        }
    }
}

bool Graph::Escape()
{
    return BFS(0);
}

bool Graph::BFS(int index)  //BFS遍历
{
    int route[4951];   //这里数组大小最好建成10000以上，因为N最多可以是100，意味着可能有N(N-1)/2个边
    int rear = -1, front = -1;
    route[++front] = 0;
    while(rear != front)
    {
        int index = ++rear;

        if(!visited[route[index]])
        {
            visited[route[index]] = 1;

            for(int i = 0; i < m_graph[route[index]].size(); ++i)
            {
                route[++front] = m_graph[route[index]][i];
            }
            if(route[index] == vertexNum + 1)
            {
                return true;
            }
        }
    }
    return false;
}

int main()
{
    int N = 0, D = 0;
    cin >> N >> D;
    Graph graph(N, D);
    cout << (graph.Escape()? "Yes":"No") << endl;

    return 0;
}

第一次提交没过因为没考虑直径15的小岛（英语渣了，题目都看不懂啊），在看网友们说了这个因素之后，改善代码，对初始位置单独处理，解决该问题，但是又蹦出新问题,就是测试点4出现段错误，段错误一般都是数组溢出导致，检查代码中的数组，找到症结所在，因为route[++front] = m_graph[route[index]][i];这句代码，我刚开始设置数组大小仅为N+1，后来考虑到可能会有完全图，任一两点之间都会有边，所以设为N(N-1)/2 = 4950,设为4951，这样这个错误就解决了。

测试结果：

测试点 结果 用时(ms) 内存(kB) 得分/满分
0 答案正确 1 180 7/7
1 答案正确 1 308 6/6
2 答案正确 1 180 3/3
3 答案正确 1 308 3/3
4 答案正确 2 308 3/3
5 答案正确 1 308 3/3