# PKU POJ 1611解题报告(并查集)

The Suspects

 Time Limit: 1000MS Memory Limit: 20000K Total Submissions: 5572 Accepted: 2660

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 4
2 1 2
5 10 13 11 12 14
2 0 1
2 99 2
200 2
1 5
5 1 2 3 4 5
1 0
0 0

Sample Output

4
1
1
我的思路：

Code 1#include<iostream> 2using namespace std; 3 4int n, m, i, j; 5int father[30005], num[30005]; 6 7void makeSet(int n) 8{ 9    for(i = 0; i < n; i++)10    {11        father[i] = i; //使用本身做根12        num[i] = 1;13    }14}15int findSet(int x)16{17    if(father[x] != x) //合并后的树的根是不变的18    {    19        father[x] = findSet(father[x]);20    }21    return father[x]; 22}2324void Union(int a, int b)25{26    int x = findSet(a);27    int y = findSet(b);28    if(x == y)29    {30        return;31    }32    if(num[x] <= num[y])33    {34        father[x] = y;35        num[y] += num[x];36    }37    else 38    {39        father[y] = x;40        num[x] += num[y];41    }42}4344int main()45{46    while(scanf("%d %d", &n, &m)!=EOF && n != 0)47    {48        makeSet(n);49        for(i = 0; i < m; i++)50        {51            int count, first, b;52            scanf("%d %d",&count, &first);53            for(j = 1; j < count; j++)54            {55                scanf("%d",&b);56                Union(first,b);57            }58        }59        printf("%d\n",num[findSet(0)]);60    }61    return 0;62}63

Code
posted @ 2009-10-11 13:07  Yx.Ac  阅读(5681)  评论(4编辑  收藏