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

标签：pat

1 边界和湖心小岛分别算一个节点，连接所有距离小于D的鳄鱼，时间复杂度O(N2)

2 判断每个连通图的节点中是否包含边界和湖心小岛，是则Yes否则No

3 冗长混乱的函数参数

#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)

标签：pat

原文地址：http://blog.csdn.net/qq_19672579/article/details/46337745