码迷,mamicode.com
首页 > 其他好文 > 详细

06-图1. List Components (25) (邻接矩阵实现)

时间:2015-05-31 09:20:28      阅读:121      评论:0      收藏:0      [点我收藏+]

标签:pat

轻松愉快的邻接矩阵,“现在空间还是问题么?”


#include <stdio.h>
#include <malloc.h>
#include <queue>
using namespace std;

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, int vertice, bool* isVisited, int N)
{
    printf("%d ", vertice);
    isVisited[vertice] = 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]])
        {
            DFS(graph, adjoinVertice[i], isVisited, N);
        }
        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++;
        }
    }
}

void MatrixComponentsSearch(int* graph, bool* isVisited, int N, int function)
{
    for (int i = 0; i < N; i++)
    {
        if (!isVisited[i])
        {
            if (function == 1)
            {
                printf("{ ");
                DFS(graph, i, isVisited, N);
                printf("}\n");
            }
            else
            {
                printf("{ ");
                BFS(graph, i, isVisited, N);
                printf("}\n");
            }
        }
    }
}

int main(void)
{
    int N;
    int E;
    scanf("%d %d", &N, &E);

    int* graph = CreateMatrixGraph(N);
    for (int i = 0; i < E; i++)
    {
        int vertical;
        int horizontal;
        scanf("%d %d", &vertical, &horizontal);

        MatrixConnect(vertical, horizontal, graph, N);
    }

    bool* isVisited = (bool*) malloc(sizeof(bool) * N);
    for (int i = 0; i < N; i++)
    {
        isVisited[i] = false;
    }
    MatrixComponentsSearch(graph, isVisited, N, 1);
    for (int i = 0; i < N; i++)
    {
        isVisited[i] = false;
    }
    MatrixComponentsSearch(graph, isVisited, N, 2);

    return 0;
}


06-图1. List Components (25) (邻接矩阵实现)

标签:pat

原文地址:http://blog.csdn.net/qq_19672579/article/details/46279323

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!