Sicily 4378. connect components in undirected graph

最新推荐文章于 2016-12-01 16:15:25 发布

原创最新推荐文章于 2016-12-01 16:15:25 发布 · 1.2k 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#components #path #graph #sun

Sicily 专栏收录该内容

121 篇文章

订阅专栏

本文介绍了一种基于多次宽度优先搜索（宽搜）的算法实现，通过实例演示了如何使用宽搜来解决特定问题，包括节点遍历及连接组件计数等。文中详细解释了宽搜的具体步骤和技巧。

仍然是宽搜，只不过是多次宽搜，输出宽搜次数即可。

Run Time: 0.01sec

Run Memory: 1168KB

Code length: 1157Bytes

Submit Time: 2011-12-24 11:55:30

// Problem#: 4378
// Submission#: 1120874
// The source code is licensed under Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License
// URI: http://creativecommons.org/licenses/by-nc-sa/3.0/
// All Copyright reserved by Informatic Lab of Sun Yat-sen University
#include <iostream>
#include <queue>
#include <cstring>
using namespace std;

int main()
{
    int n, m;
    int i, j;
    queue<int> q;
    int count, set;
    bool visited[ 1001 ];
    bool path[ 1001 ][ 1001 ];

    memset( visited, false, sizeof( visited ) );
    memset( path, false, sizeof( path ) );
    cin >> n >> m;
    while ( m-- ) {
        cin >> i >> j;
        path[ i ][ j ] = true;
        path[ j ][ i ] = true;
    }

    count = 0;
    set = 0;
    while ( count != n ) {
        for ( i = 1; i <= n; i++ ) {
            if ( visited[ i ] == false ) {
                q.push( i );
                visited[ i ] = true;
                count++;
                set++;
                break;
            }
        }

        while ( !q.empty() ) {
            i = q.front();
            for ( j = 1; j <= n; j++ ) {
                if ( path[ i ][ j ] && !visited[ j ] ) {
                    q.push( j );
                    visited[ j ] = true;
                    count++;
                }
            }
            q.pop();
        }
    }
    cout << set << endl;

    return 0;

}