poj2524-Ubiquitous Religions

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C - Ubiquitous Religions

Time Limit: 5000 MS Memory Limit: 65536 KB

64-bit integer IO format: %I64d , %I64u Java class name: Main

DescriptionThere are so many different religions in the world today that it is difficult to keep track of them all. You are interested in finding out how many different religions students in your university believe in. 

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.InputThe input consists of a number of cases. Each case starts with a line specifying the integers n and m. The next m lines each consists of two integers i and j, specifying that students i and j believe in the same religion. The students are numbered 1 to n. The end of input is specified by a line in which n = m = 0.OutputFor each test case, print on a single line the case number (starting with 1) followed by the maximum number of different religions that the students in the university believe in.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

HintHuge input, scanf is recommended. 並查集模板題簡單變新，前邊這種是完全沒有最佳化的，所以時間巨長！！      錯了４發，ＰＥ，這個—_—!

//3469 ms 544 KB C++ 1295 B#include <cstdio>using namespace std;const int maxn = 50005;int father[maxn];int mark[maxn];void init(int n){    for(int i = 1; i <= n; ++i)    {        father[i] = i;        mark[i] = 0;    }}int serch(int x){    if(father[x] == x)        return x;    return father[x] = serch(father[x]);}void join(int x, int y){    int fx = serch(x), fy = serch(y);    if(fx != fy)        father[fx] = fy;}int main(){    int n, m, a, b;    int casee = 0;    while(scanf("%d %d", &n, &m)!= EOF && (n || m))    {        casee++;        init(n);        for(int i = 1; i <= m; ++i)        {            scanf("%d %d", &a, &b);            join(a, b);        }        int res = 0;        for(int i = 1; i <= n; ++i)        {            if(!mark[i])            {                mark[i] = 5;                for(int j = 1; j <= n; ++j)                {                    if(!mark[j] && (serch(i) == serch(j)))                    {                        mark[j] = 1;                    }                }            }        }        for(int i = 1; i <= n; ++i)        {            if(mark[i] == 5)                res++;        }        printf("Case %d: %d\n", casee, res);    }    return 0;}

　　

 採用路徑壓縮後能快點，路徑壓縮只需改一句話。當然還有其他最佳化，還在學習中。。。 

///Accepted3469 ms544 KBC++1295 Bint serch(int x){    if(father[x] == x)        return x;    return father[x] = serch(father[x]);　　　///以前為 return  serch(father[x]);　}