并查集学习篇

    因为虚拟oj上挂了一道带权并查集的题，所以我只有开足马力学习并查集知识，还好网上关于这方面的内容足够详细（正好能满足自己的自学），在POJ有两道基础好题，先贴出，并附上自己敲代码时碰到的问题，

A

The Suspects

Time Limit: 1000MS Memory Limit: 20000KTotal Submissions: 23792 Accepted: 11614

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

Source

Asia Kaohsiung 2003

代码如下

#include<cstdio>#include<iostream>using namespace std;int n,m;int fa[30000],rank[30000],num[30000];void init(int n){ for(int i=0;i<n i="" fa="" i="" i="" rank="" i="" 0="" rank="" num="" ac="" num="" i="" 1="" int="" find_set="" int="" x="" if="" x="" fa="" x="" if="" while="" while="" fa="" x="" find_set="" fa="" x="" return="" fa="" x="" void="" union="" int="" x="" int="" y="" x="find_set(x);" y="find_set(y);" if="" x="=y)return" if="" rank="" x="">rank[y]) { fa[y]=x; num[x]=num[x]+num[y]; } else { fa[x]=y; if(rank[x]==rank[y])rank[y]++; num[y]+=num[x]; }}int main(){ int a,b,c; while(cin>>n>>m) { if(n==0&&m==0)break; init(n); while(m--) { cin>>a>>b; for(int i=1;i<a;i++) { // cout<<i<<endl cin="">>c; Union(b,c); // cout<<i<<endl; } } cout<<num[find_set(0)]<<endl; //注意这里不是rank[0]!! } return 0;}</endl></n></iostream></cstdio>

B

Ubiquitous Religions

Time Limit: 5000MS Memory Limit: 65536KTotal Submissions: 25520 Accepted: 12606

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 91 21 31 41 51 61 71 81 91 1010 42 34 54 85 80 0

Sample Output

Case 1: 1Case 2: 7

Hint

Huge input, scanf is recommended.

Source

Alberta Collegiate Programming Contest 2003.10.18

  

现在先附上自己的wa代码，原因未知（权当参考，日后待查）

#include<iostream>using namespace std;#define nn 50005int fa[nn],rank[nn];void init(int n){ for(int i=1;i<=n;i++) { fa[i]=i; rank[i]=1; }}int find_set(int x){ if(x!=fa[x]) fa[x]=find_set(fa[x]); else return fa[x];}void Union(int x,int y){ int x1=x,y1=y; x=find_set(x); y=find_set(y); if(x==y)return; if(rank[x]>=rank[y]) { fa[y]=x; rank[x]=rank[x]+rank[y]; rank[y1]=0; } else { fa[x]=y; rank[y]=rank[x]+rank[y]; rank[x1]=0; }}int main(){ int n,m,a,b,num=0,j=0; while(cin>>n>>m) { if(n==0&&m==0)break; init(n); for(int i=0;i<m i="" cin="">>a>>b; Union(a,b); } for(int i=1;i<=n;i++) { if(rank[i]>0)num++; } j++; cout<<"Case "<<j<<": "<<num<<endl; num=0; } return 0;}</m></iostream>

后面贴出的是看了网上题解后自己敲出的ac代码

#include<iostream>#include<cstdio>using namespace std;#define nn 50005int fa[nn],rank[nn],num[nn],ans;void init(int n){ for(int i=1;i<=n;i++) { fa[i]=i; rank[i]=0; // num[i]=1; }}int find_set(int x){ if(x!=fa[x]) fa[x]=find_set(fa[x]); else return fa[x];}void Union(int x,int y){ // int x1=x,y1=y; x=find_set(x); y=find_set(y); if(x==y)return; ans--; //此处是点金之笔 if(rank[x]>rank[y]) { fa[y]=x; // num[x]+=num[y]; } else { fa[x]=y; if(rank[x]==rank[y])rank[y]++; // num[y]+=num[x]; }}int main(){ int n,m,a,b,j=0; while(cin>>n>>m) { if(n==0&&m==0)break; init(n); ans=n; for(int i=0;i<m i="" cin="">>a>>b; Union(a,b); } /* for(int i=1;i<=n;i++) { cout<<num[find_set(i)]<<endl if="" rank="" fa="" i="">0)num++; }*/ j++; cout<<"Case "<<j<<": "<<ans<<endl; } return 0;}</endl></m></cstdio></iostream>

哈哈 今天又学了一手好并查集，明天就要回家了，希望在回家前能成功ac下那道带权并查集的题目，，晚安 我可爱的世界