poj–1611–並查集（路徑壓縮運用）

The Suspects

Time Limit: 1000MS		Memory Limit: 20000K
Total Submissions: 18782		Accepted: 9084

Description

Severe acute respiratory syndrome (SARS), an atypical pneumonia of unknown aetiology, was recognized as a global threat in mid-March 2003. To minimize transmission to others, the best strategy is to separate the suspects from others.
In the Not-Spreading-Your-Sickness University (NSYSU), there are many student groups. Students in the same group intercommunicate with each other frequently, and a student may join several groups. To prevent the possible transmissions of SARS, the NSYSU collects
the member lists of all student groups, and makes the following rule in their standard operation procedure (SOP).
Once a member in a group is a suspect, all members in the group are suspects.
However, they find that it is not easy to identify all the suspects when a student is recognized as a suspect. Your job is to write a program which finds all the suspects.

Input

The input file contains several cases. Each test case begins with two integers n and m in a line, where n is the number of students, and m is the number of groups. You may assume that 0 < n <= 30000 and 0 <= m <= 500. Every student
is numbered by a unique integer between 0 and n−1, and initially student 0 is recognized as a suspect in all the cases. This line is followed by m member lists of the groups, one line per group. Each line begins with an integer k by itself representing the
number of members in the group. Following the number of members, there are k integers representing the students in this group. All the integers in a line are separated by at least one space.
A case with n = 0 and m = 0 indicates the end of the input, and need not be processed.

Output

For each case, output the number of suspects in one line.

Sample Input

100 42 1 25 10 13 11 12 142 0 12 99 2200 21 55 1 2 3 4 51 00 0

Sample Output

411

#include <iostream>#include<stdio.h>#include<stdlib.h>using namespace std;const int MAXN = 30500; /*必須開大點/int pa[MAXN];    /*p[x】表示x的父節點int rank[MAXN];    /*rank是x的高度的一個上界/int num[MAXN];/*num[]//存的集合中的元素個數void make_init(int x){    pa[x] = x;    rank[x] = 0;    num[x] = 1;}int find_set(int x){    int r = x, temp;    while(pa[r] != r) r = pa[r];    while(x != r)    {        temp = pa[x];        pa[x] = r;        x = temp;    }    return x;    //if(x != pa[x]) //遞迴壓縮。不太好，容易溢出棧

    //    pa[x] = find_set(pa[x]);    //return pa[x];}void union_set(int x, int y){    x = find_set(x);    y = find_set(y);    if(x == y)return ;    if(rank[x] > rank[y])/*讓rank比較高的作為父節點    {        pa[y] = x;        num[x] += num[y];    }    else     {        pa[x] = y;        if(rank[x] == rank[y])            rank[y]++;        num[y] += num[x];    }}//answer to 1611 int main(){    int n, m, x, y, i, t, j;    while(scanf("%d%d", &n, &m))    {        if(m==n && n == 0) break;        if(m == 0)        {        printf("1\n");            continue;        }        for(i = 0; i < n; i++){          make_init(i);        }            for(i = 0; i < m; i++)        {            scanf("%d", &t);            scanf("%d", &x);            for(j = 1; j < t; j++){                scanf("%d", &y);                union_set(x, y);                x = y;            }        }        x = find_set(0);                 printf("%d\n",num[x]);    }    return 0;}

 

 

 

 

這是樸素尋找的代碼,適合資料量不大的情況：

int findx(int x)
{
int r=x;
while(parent[r] !=r)
        r=parent[r];
return r;
}

   

    下面是採用路徑壓縮的方法尋找元素：

int find(int x)       //尋找x元素所在的集合,回溯時壓縮路徑
{
if (x != parent[x])
    {
        parent[x] = find(parent[x]);     //回溯時的壓縮路徑
    }         //從x結點搜尋到祖先結點所經過的結點都指向該祖先結點
    return parent[x];
}

    

    上面是一採用遞迴的方式壓縮路徑， 但是，遞迴壓縮路徑可能會造成溢出棧，我曾經因為這個RE了n次，下面我們說一下非遞迴方式進行的路徑壓縮：

int find(int x)
{
int k, j, r;
    r = x;
while(r != parent[r])     //尋找跟節點
        r = parent[r];      //找到跟節點，用r記錄下
    k = x;        
while(k != r)             //非遞迴路徑壓縮操作
    {
        j = parent[k];         //用j暫存parent[k]的父節點
        parent[k] = r;        //parent[x]指向跟節點
        k = j;                    //k移到父節點
    }
return r;         //返回根節點的值            
}