poj 2524 Ubiquitous Religions
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Description
You know that there are n students in your university (0 < n <= 50000). It is infeasible for you to ask every student their religious beliefs. Furthermore, many students are not comfortable expressing their beliefs. One way to avoid these problems is to ask m (0 <= m <= n(n-1)/2) pairs of students and ask them whether they believe in the same religion (e.g. they may know if they both attend the same church). From this data, you may not know what each person believes in, but you can get an idea of the upper bound of how many different religions can be possibly represented on campus. You may assume that each student subscribes to at most one religion.
Input
Output
Sample Input
10 91 21 31 41 51 61 71 81 91 1010 42 34 54 85 80 0
Sample Output
Case 1: 1Case 2: 7
分析:每访问一对学生就把他们两个所在集合合并,最后查有几个不同集合就可以了。
代码
const maxn=50000;var f:array[0..maxn] of boolean; p:array[0..maxn] of longint; i,j,n,m,ans,x,y:longint;function find(x:longint):longint;begin if p[x]=x then exit(x); p[x]:=find(p[x]); exit(p[x]);end;begin readln(n,m); while (n<>0) or (m<>0) do begin inc(j); for i:=1 to n do p[i]:=i; ans:=0; fillchar(f,sizeof(f),false); for i:=1 to m do begin readln(x,y); p[find(x)]:=find(y); end; for i:=1 to n do if not f[find(i)] then begin f[find(i)]:=true; inc(ans); end; writeln('Case ',j,': ',ans); readln(n,m); end;end.
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