Get started with a check set

I was looking at "Aha!" Algorithm "This book completed and find the introduction of the set, I would like to cite another chestnut, but ... The chestnut in the book is very apt.

First introduce a problem: known to have 10 bandits, the police need a little bit of trace and finally dug up their respective behind the gang Yiguoduan, after a period of time the investigation of the police got 9 exact clues, respectively, can explain the two bandits of the relationship. So how many gangs will be knocked out of this operation altogether?

Enter the following data:

Follow the first line to enter the number of people n, the number of threads m, the next m line input clues, each line of clues such as 1 2 for Bandits 1th and 2nd is a gang. A set of data is now available for follow-up and program testing:

<span style= "FONT-SIZE:18PX;" >10 (4</span>)

How many outputs should this set of data be? Let's make a little analysis.

1: First to apply for a one-dimensional array f, with subscript 1~10 to mark the 10 bandits, each element corresponding to the number of each of the bandits to represent the boss who.

2: Initialization, first we assume that each bandit "fragmented", that is, F[1]=1, f[2]=2 ... f[i] = i;

3: Start processing Clues, first clue 1 2, you can find 1th and 2nd is a gang, then in the end is to make 1th surrender 2nd or 2nd to surrender 1th? Here we first follow the "left" rule, let number 2nd, that is, a f[2] = 1; At this point the F array becomes

1 1 3 4 5 6 7 8 9 10

Then start processing the second clue, 3 4, in the same way according to the "left law" f[4] = 3; At this point the F array becomes:

1 1 3 3 5 6 7 8 9 10

Continue the third Clue 5 2, known f[5] = 5, description of the boss of 5th is himself, for 2nd f[2] = 1; Note 2nd is the boss of 1th, at this time still in accordance with the "left" rule, let 2nd surrender 5th is f[2] = 5; Then the robber of the number 1th will quit. Why the Gun my Man (f[2] =1)? When there is contradiction, how to do??? Introduce another law at this time-----> The thief first to seize the king!! Directly find 2nd boss talk, let 2nd number of Boss 1th surrender 5th is f[1]=5, conflict resolution!! The third thread is finished, at this point the F array is:

5 5 3 3 5 6 7 8 9 10

Here's the rule of the left and the catch thief. The 6 clues after the king's rule can be handled smoothly, not in the process of repeating, the F array after each processing is given directly

5 5 3 3 5 3 7 8 9 10

5 5 5 3 5 5 78 9 10

5 5 5 3 5 5 8 8 9 10

5 5 5 3 5 5 9 9 9 10

The ultimate gang relationship is clearly present:

2 1 3 4 6 boss is number 5th

Number 10th, be your own boss.

8 7 boss is number 9th.

So a total of three gangs! Problem solved satisfactorily,

PS: The detailed process can refer to the "Aha!" Algorithm "200 page, the following gives the problem solving code:

<span style= "FONT-SIZE:18PX;"   > #include <cstdio> #include <iostream>using namespace std; int f[1000];            Define individual elements int n,m;           The total number of N and the number of known relationships int x, y;    A set of relationships such as 1 2 for 1th number 2nd for a gang of void init ()//initialization {for (int i = 1; i<=n; i++) {f[i] = i;    }}int getf (int v)//Keep looking for your host {if (f[v] = = v) return v;        else {F[v] = GETF (F[v]);    return F[V];    }}void merge (int v,int u)//merge function {int t1,t2;    T1 = GETF (v);    t2 = GETF (u);    if (t1! = t2)//According to the principle of the left priority, let T2 submit T1 {f[t2] = T1;        }}int Main () {while (scanf ("%d%d", &n,&m)!=eof) {init ();            for (int i = 0; i<m; i++) {scanf ("%d%d", &x,&y);        Merge (x, y);      } int sum = 0;        Statistic gang number sum for (int i = 1; i<=n; i++) {if (f[i] = = i) sum++;    } printf ("A total of%d gang \ n", sum); } return 0;} </span>

Get started with a check set