POJ-1611—The Suspects

客官°小女子只卖身不卖艺 2022-03-09 01:10 177阅读 0赞

# [1611–The Suspects][1611_The Suspects] #

# 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

### 并查集 ###

经典的并查集应用，把有交集的组合并为一个组，最后查找与节点0的根节点相同的节点。

#include <vector>
    #include <iostream>
    using namespace std;
    int n,m,k,t,f,a,b;
    vector<int> p(30001),rands(30001,0); //rands数组起一个优先级作用，不代表根节点下子节点的数量
    int find(int x) //查找
    { 
        if(x == p[x])
            return x;
        else return p[x] = find(p[x]);
    }
    void merig(int x,int y) //压缩
    { 
        a = find(x); //查找x的根节点
        b = find(y); //查找y的根节点
        if(a == b)
            return;
        if(rands[a]>rands[b]) //如果a的子节点数比b的多，就把b归为a的子节点
        { 
            p[b] = a;
        }
        else                  //否则，把a归为b的子节点
        { 
            p[a] = b;
            if(rands[a] == rands[b])
                rands[b]++; 
        }
    }
    int main()
    { 
        while(cin>>m>>n&&(m||n)) //m个学生，n个组
        { 
            for(int i=0;i<m;i++) //初始化每个节点的根结点为自己
            { 
                p[i]=i; //表示i的根节点为p[i]
            }
            for(int i=0;i<n;i++)
            { 
                cin>>k; //每个组k个学生
                if(k>=1)
                    cin>>f;
                for(int j=1;j<k;j++)
                { 
                    cin>>t; 
                    merig(f,t); //重新设定根节点，
                }
            }
            int sum = 1;
            for(int i=1;i<m;i++)
            { 
                // if(find(i)==find(0))
                if(p[i] == p[0]) //因为p[i]存的就是i的根节点，所以不必再调用find函数
                { 
                    sum++;
                }
            }
            cout<<sum<<endl;
        }
        return 0;
    }

### 优化 ###

可以将rands数组改为保存根节点下子节点的数量，每次变换根节点时就更新，最后直接输出根节点的rands即可

#include <cstdio>
    #include <vector>
    #include <iostream>
    #include <cstring>
    using namespace std;
    int n,m,k,t,f,a,b;
    vector<int> p(30001),rands(30001,1);
    int find(int x)
    { 
        if(x == p[x])
            return x;
        else return p[x] = find(p[x]);
    }
    void merig(int x,int y) //连接
    { 
        a = find(x);
        b = find(y);
        if(a == b)
            return;
        if(rands[a]>rands[b])
        { 
            p[b] = a;
            rands[a]++;
        }
        else
        { 
            p[a] = b;
            //if(rands[a] == rands[b])
            rands[b]++;
        }
    }
    int main()
    { 
        while(cin>>m>>n&&(m||n)) //m个学生，n个组
        { 
            for(int i=0;i<m;i++)
            { 
                p[i]=i;
            }
            for(int i=0;i<n;i++)
            { 
                cin>>k; //每个组k个学生
                if(k>=1)
                    cin>>f;
                for(int j=1;j<k;j++)
                { 
                    cin>>t; 
                    merig(f,t);
                }
            }
            cout<<rands[p[0]]<<endl;
        }
        return 0;
    }

[1611_The Suspects]: http://poj.org/problem?id=1611