算法-动态连通性

动态连通性直接听起来会稍微绕口一点,简单的说就是输入一列整数对,其中每个整数都表示某种类型的对象,假设输入的的整数对是p和q,我们可以理解p和q是相连的,假设相连是一种等价关系,一般具有三种特性自反性,对称性,传递性,根据上面的特性,如果整数对不存在某种等价关系,那么直接输出,如果存在就不输出。简单的举几个例子,计算机网络中判断两个计算机是否需要建立新的连接通信,如果可以通过一个或某几个节点能通信,那么我们就不需要建立新的连接,数学中可以将p和q看成集合,跑题了,看下如何实现的吧,四种方式依次渐进:

Quick-Find算法

p和q做网络上相连可以看成连接,单独的看p和q可以看成触点,可以判断是否存在p和q或者pq之间的等价连接:

@interface DynamicUnion : NSObject



@property (strong,nonatomic) NSMutableArray  *ids;//存储每个触点对应的值


@property (assign,nonatomic) NSInteger  count;//已经连通的连接的数量


-(BOOL)connected:(NSInteger)a secondNumber:(NSInteger)b;//是否已经存在连接或者等价的连接

-(NSInteger)find:(NSInteger)index;//取出触点的值

-(void)dynamicUnion:(NSInteger)a secondNumber:(NSInteger)b;//a,b之间建立一个连接

-(void)initData:(NSInteger)count;//初始化触点的数量


@end

具体实现代码: 

-(void)initData:(NSInteger)count{
    self.count=count;
    self.ids=[[NSMutableArray alloc]initWithCapacity:count];
    for (NSInteger i=0; i<count; i++) {
        [self.ids addObject:[NSNumber numberWithInteger:i]];
    }
}

//http://www.cnblogs.com/xiaofeixiang
-(BOOL)connected:(NSInteger)a secondNumber:(NSInteger)b{
    return [self find:a]==[self find:b];
}

-(NSInteger)find:(NSInteger)index{
    return [[self.ids objectAtIndex:index] integerValue];
}

-(void)dynamicUnion:(NSInteger)a secondNumber:(NSInteger)b{
    NSInteger aID=[self find:a];
    NSInteger bID=[self find:b];
    if (aID==bID) {
        return;
    }
    for (NSInteger i=0;i<[self.ids count]; i++) {
        if ([[self.ids objectAtIndex:i] integerValue]==aID) {
            self.ids[i]=[NSNumber numberWithInteger:bID];
        }
    }
    self.count=self.count-1;
}

Quick-Union算法

 Quick-Find在Union的过程中,每次都会遍历数组一次,这样有损性能,还是用ids数组存每个触点的值,不过每个值存的意义不一样,我们可以通过ids中的值存储触点的父级,作为一棵树存在,我们就不用遍历ids数组,简单的讲就是4,3的时候ids[4]存值的时候存的是3,这样最后会只要判断根节点就可以。其他方法不用变,我们只需要改变find和dynamicUnion方法即可。

-(NSInteger)find:(NSInteger)index{
    while (index!=[[self.ids objectAtIndex:index] integerValue]) {
        index=[[self.ids objectAtIndex:index] integerValue];
    }
    return index;
}
//http://www.cnblogs.com/xiaofeixiang
-(void)dynamicUnion:(NSInteger)a secondNumber:(NSInteger)b{
    NSInteger aRoot=[self find:a];
    NSInteger bRoot=[self find:b];
    if (aRoot==bRoot) {
        return;
    }
    self.ids[aRoot]=[NSNumber numberWithInteger:bRoot];
    self.count--;
 }

加权Quick-Union算法

Quick-Union可能会出现一种情况,如果以树的形式去展示的话,最终可能会出现大树的父级别是小树的情况,因为我们需要通过一个权重值,避免出现这种情况,不过我们需要回顾一个树的基础概念。一棵树的大小是它的节点的数量,树中的一个节点的深度是它到根节点的路径上的链接数。输的高度是它的所有节点中的最大深度。加权能保每次find、connected和union都是lgn级别。

//http://www.cnblogs.com/xiaofeixiang
@interface DynamicUnionWeight : NSObject


@property (strong,nonatomic) NSMutableArray  *ids;//存储每个触点对应的值
@property (strong,nonatomic) NSMutableArray  *weightArr;//各个根节点对ing的分量的大小


@property (assign,nonatomic) NSInteger  count;//已经连通的连接的数量


-(BOOL)connected:(NSInteger)a secondNumber:(NSInteger)b;//是否已经存在连接或者等价的连接

-(NSInteger)find:(NSInteger)index;//取出触点的值

-(void)dynamicUnion:(NSInteger)a secondNumber:(NSInteger)b;//a,b之间建立一个连接

-(void)initData:(NSInteger)count;//初始化触点的数量

@end 

具体实现:

-(void)initData:(NSInteger)count{
    self.count=count;
    self.ids=[[NSMutableArray alloc]initWithCapacity:count];
    self.weightArr=[[NSMutableArray alloc]initWithCapacity:count];
  

    for (NSInteger i=0; i<count; i++) {
        [self.ids addObject:[NSNumber numberWithInteger:i]];
        [self.weightArr addObject:[NSNumber numberWithInteger:1]];
    }
    
}

//http://www.cnblogs.com/xiaofeixiang
-(BOOL)connected:(NSInteger)a secondNumber:(NSInteger)b{
    return [self find:a]==[self find:b];
}



-(NSInteger)find:(NSInteger)index{
    while (index!=[[self.ids objectAtIndex:index] integerValue]) {
        index=[[self.ids objectAtIndex:index] integerValue];
    }
    return index;
}

-(void)dynamicUnion:(NSInteger)a secondNumber:(NSInteger)b{
    NSInteger i=[self find:a];
    NSInteger j=[self find:b];
    if (i==j) {
        return;
    }
    NSInteger  weightA=[[self.weightArr objectAtIndex:i] integerValue];
    NSInteger  weightB=[[self.weightArr objectAtIndex:j] integerValue];
 
    if (weightA<weightB) {
        self.ids[i]=[NSNumber numberWithInteger:j];
        self.weightArr[j]=[NSNumber numberWithInteger:weightA+weightB];
    }else{
        self.ids[j]=[NSNumber numberWithInteger:i];
        self.weightArr[i]=[NSNumber numberWithInteger:weightA+weightB];
    }
    self.count--;
}

路径压缩算法

路径压缩会保证union都接近于1,这个实现只需要加权Quick-Union的算法稍微改动一下,改变一下find即可:

-(NSInteger)find:(NSInteger)index{
    while (index!=[[self.ids objectAtIndex:index] integerValue]) {
         self.ids[index]=self.ids[[self.ids[index] integerValue]];
        index=[[self.ids objectAtIndex:index] integerValue];
    }
    return index;
}
//http://www.cnblogs.com/xiaofeixiang
-(void)dynamicUnion:(NSInteger)a secondNumber:(NSInteger)b{
    NSInteger i=[self find:a];
    NSInteger j=[self find:b];
    if (i==j) {
        return;
    }
    NSInteger  weightA=[[self.weightArr objectAtIndex:i] integerValue];
    NSInteger  weightB=[[self.weightArr objectAtIndex:j] integerValue];
 
    if (weightA<weightB) {
        self.ids[i]=[NSNumber numberWithInteger:j];
        self.weightArr[j]=[NSNumber numberWithInteger:weightA+weightB];
    }else{
        self.ids[j]=[NSNumber numberWithInteger:i];
        self.weightArr[i]=[NSNumber numberWithInteger:weightA+weightB];
    }
    self.count--;
}

 模拟测试:

 DynamicUnion  *dynamicUnion=[[DynamicUnion alloc]init];
        [dynamicUnion initData:10];
        NSMutableArray *dataSource=[[NSMutableArray alloc]initWithCapacity:10];
        [dataSource addObject:@"4,3"];
        [dataSource addObject:@"3,8"];
        [dataSource addObject:@"6,5"];
        [dataSource addObject:@"9,4"];
        [dataSource addObject:@"2,1"];
        [dataSource addObject:@"8,9"];
        [dataSource addObject:@"5,0"];
        [dataSource addObject:@"7,2"];
        [dataSource addObject:@"6,1"];
        [dataSource addObject:@"1,0"];
        
        for (NSInteger i=0; i<[dataSource count]; i++) {
            NSString *node=[dataSource objectAtIndex:i];
            NSInteger a=[[node substringWithRange:NSMakeRange(0, 1)] integerValue];
            NSInteger b=[[node substringWithRange:NSMakeRange(2, 1)] integerValue];
            if ([dynamicUnion connected:a secondNumber:b]) {
                continue;
            }
            [dynamicUnion dynamicUnion:a secondNumber:b];
            NSLog(@"%ld---%ld",a,b);
        }
        NSLog(@"%ld已存在连接",dynamicUnion.count);
        NSLog(@"iOS技术交流群:228407086");
        NSLog(@"原文地址:http://www.cnblogs.com/xiaofeixiang");

结果如下:

posted @ 2015-06-15 07:32  Fly_Elephant  阅读(1618)  评论(0编辑  收藏  举报