POJ 3660Cow Contest (query set + topological sorting)

POJ 3660Cow Contest (query set + topological sorting)
Cow Contest

Time Limit:1000 MS		Memory Limit:65536 K
Total Submissions:7567		Accepted:4206

Description

N(1 ≤N≤ 100) cows, conveniently numbered 1 ..N, Are participating in a programming contest. As we all know, some cows code better than others. Each cow has a certain constant skill rating that is unique among the competitors.

The contest is conducting CTED in several head-to-head rounds, each between two cows. If cowAHas a greater skill level than cowB(1 ≤A≤N; 1 ≤B≤N;A=B), Then cowAWill always beat cowB.

Farmer John is trying to rank the cows by skill level. Given a list the resultsM(1 ≤M≤ 4,500) two-cow rounds, determine the number of cows whose ranks can be precisely determined from the results. It is guaranteed that the results of the rounds will not be contradic.pdf.

Input

* Line 1: Two space-separated integers:NAndM
* Lines 2 ..M+ 1: Each line contains two space-separated integers that describe the competitors and results (the first integer,A, Is the winner) of a single round of competition:AAndB

Output

* Line 1: A single integer representing the number of cows whose ranks can be determined
　

Sample Input

5 54 34 23 21 22 5

Sample Output

2

Source

Question: There are n cows, numbered 1 ~ N. M relationships are given: a B, which means A is better than B. The question is how many cows can be uniquely identified (that is, the relationship between this ox and n-1 ox is uniquely identified ).

Problem-solving: If a cow can be uniquely identified, then all points must be a connected block, so it may be used to query the set for determination. The next step is the topological sorting. For more information, see the code.

 

# Include
 
  
# Include
  
   
Const int n= 105; bool mapt [N] [N], path [N] [N]; int N, in [n], father [N]; void init () {for (int I = 1; I <= n; I ++) {father [I] = I; in [I] = 0; for (int j = 1; j <= n; j ++) mapt [I] [j] = path [I] [j] = 0; path [I] [I] = 1 ;}} int findroot (int x) {if (x! = Father [x]) father [x] = findroot (father [x]); return father [x];} void setroot (int x, int y) {x = findroot (x); y = findroot (y); father [x] = y;} int tope () {int a [N], k = 0, l = 0, ans = 0; for (int I = 1; I <= n; I ++) if (in [I] = 0) a [k ++] = I; while (l
   
    
0) {init (); while (m --) {scanf ("% d", & a, & B); setroot (a, B ); if (mapt [a] [B] = 0) mapt [a] [B] = 1, in [B] ++;} int k = 0; for (int I = 1; I <= n; I ++) if (father [I] = I) k ++; if (k> 1) // It indicates that all vertices are not in a connected block and all vertices cannot be determined printf ("0 \ n"); else printf ("% d \ n ", tope ());}}