poj 2524 Ubiquitous Religions(*信仰）

Ubiquitous Religions

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 30666		Accepted: 14860

Description

There 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.

Input

The 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.

Output

For 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 9
1 2
1 3
1 4
1 5
1 6
1 7
1 8
1 9
1 10
10 4
2 3
4 5
4 8
5 8
0 0

Sample Output

Case 1: 1
Case 2: 7

Hint

Huge input, scanf is recommended.

 

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <fstream>
using namespace std;
int f[50005];

int find(int x) {
    if (x != f[x])
        f[x] = find(f[x]);
    return f[x];
}

void Union(int a, int b) {
    int f1 = find(a);
    int f2 = find(b);
    if (f1 != f2)
        f[f2] = f1;
}

int main() {
    //ifstream cin("aaa.txt");
    int n, m, test = 1, sum;

    while(scanf("%d%d", &n, &m)){
        if (n == 0 && m == 0)
            break;
        memset(f, 0, sizeof(f));

        for(int i = 1; i <= n; i++) {
            f[i] = i;
        }

        sum = n;

        for(int i = 1; i <= m; i++) {
            int a, b;
            scanf("%d%d", &a, &b);
            if(find(a) != find(b)){
                Union(a, b);
                sum--;
            }
        }

        printf("Case %d: %d
", test++, sum);

    }
    //system("pause");
    return 0;
}

#include <stdio.h>
#include <iostream>
using namespace std;

const int MAXN = 50005; /*结点数目上线*/
int pa[MAXN];    /*p[x]表示x的父节点*/
int rank1[MAXN];    /*rank[x]是x的高度的一个上界*/
int n, ans; 

void make_set(int x)
{/*创建一个单元集*/
    pa[x] = x;
    rank1[x] = 0;
}

int find_set(int x)
{/*带路径压缩的查找*/
    if(x != pa[x])
        pa[x] = find_set(pa[x]);
    return pa[x];
}

/*按秩合并x，y所在的集合*/
void union_set(int x, int y)
{
    x = find_set(x);
    y = find_set(y);
    if(x == y)return ;
    ans--;    //统计
    if(rank1[x] > rank1[y])/*让rank比较高的作为父结点*/
    {
        pa[y] = x;
    }
    else 
    {
        pa[x] = y;
        if(rank1[x] == rank1[y])
            rank1[y]++;
    }
}
//answer to 2524
int main()
{
    int m, i, j = 1, x, y;
    while(scanf("%d%d", &n, &m))
    {
        if(n == m && m == 0) break;
        for(i = 1; i <= n; i++)
            make_set(i);
        ans = n;
        for(i = 0; i < m; i++)
        {
            scanf("%d%d", &x, &y);
            union_set(x, y);
        }
        printf("Case %d: %d
", j, ans);
        j++;
    }
    return 0;
}