题目来源：http://acm.hdu.edu.cn/showproblem.php?pid=2473
Junk-Mail Filter
Time Limit: 15000/8000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5343 Accepted Submission(s): 1702
Problem Description
Recognizing junk mails is a tough task. The method used here consists of two steps:
1) Extract the common characteristics from the incoming email.
2) Use a filter matching the set of common characteristics extracted to determine whether the email is a spam.
We want to extract the set of common characteristics from the N sample junk emails available at the moment, and thus having a handy data-analyzing tool would be helpful. The tool should support the following kinds of operations:
a) “M X Y”, meaning that we think that the characteristics of spam X and Y are the same. Note that the relationship defined here is transitive, so
relationships (other than the one between X and Y) need to be created if they are not present at the moment.
b) “S X”, meaning that we think spam X had been misidentified. Your tool should remove all relationships that spam X has when this command is received; after that, spam X will become an isolated node in the relationship graph.
Initially no relationships exist between any pair of the junk emails, so the number of distinct characteristics at that time is N.
Please help us keep track of any necessary information to solve our problem.
1) Extract the common characteristics from the incoming email.
2) Use a filter matching the set of common characteristics extracted to determine whether the email is a spam.
We want to extract the set of common characteristics from the N sample junk emails available at the moment, and thus having a handy data-analyzing tool would be helpful. The tool should support the following kinds of operations:
a) “M X Y”, meaning that we think that the characteristics of spam X and Y are the same. Note that the relationship defined here is transitive, so
relationships (other than the one between X and Y) need to be created if they are not present at the moment.
b) “S X”, meaning that we think spam X had been misidentified. Your tool should remove all relationships that spam X has when this command is received; after that, spam X will become an isolated node in the relationship graph.
Initially no relationships exist between any pair of the junk emails, so the number of distinct characteristics at that time is N.
Please help us keep track of any necessary information to solve our problem.
Input
There are multiple test cases in the input file.
Each test case starts with two integers, N and M (1 ≤ N ≤ 105 , 1 ≤ M ≤ 106), the number of email samples and the number of operations. M lines follow, each line is one of the two formats described above.
Two successive test cases are separated by a blank line. A case with N = 0 and M = 0 indicates the end of the input file, and should not be processed by your program.
Each test case starts with two integers, N and M (1 ≤ N ≤ 105 , 1 ≤ M ≤ 106), the number of email samples and the number of operations. M lines follow, each line is one of the two formats described above.
Two successive test cases are separated by a blank line. A case with N = 0 and M = 0 indicates the end of the input file, and should not be processed by your program.
Output
For each test case, please print a single integer, the number of distinct common characteristics, to the console. Follow the format as indicated in the sample below.
Sample Input
5 6
M 0 1
M 1 2
M 1 3
S 1
M 1 2
S 3
3 1
M 1 2
0 0
Sample Output
Case #1: 3
Case #2: 2
分析：
本题考查的是并查集的删除.
在删除结点的时候要注意这样的问题: 并查集的结构是一棵树,当把某一个结点删去(将其父结点置为自己)时,它的子结点的根就会丢失.因此,在删除某结点的时候,要确定它的所有子结点都已经并到根上了.
解决方法: 为每一个结点加一个虚根,这样每个结点都是叶子结点.插入结点时,把它们都并到虚根的集合中.删除结点时,只要把它的父结点置为一个无用的虚根，找个点顶替要删除的点，就行了。
代码如下：
#include <cstdlib> #include <cstring> #include <algorithm> #include <cstdio> #include <cmath> #include <iostream> #include <vector> #include<set> using namespace std; typedef long long ll; #define N 1100010 int extra; struct Node { int father; int replace; }; Node node[N]; //为每一个节点加一个虚根，这样每个节点都是叶子节点， void init(int n) // n为节点个数 { int i; for(i=0;i<n;i++) // 下标从0开始 { node[i].father=i; node[i].replace=i; } extra=n; // 虚拟节点 } int find(int x) { return node[x].father!=x?node[x].father=find(node[x].father):node[x].father; } //插入节点时，把他们都并到虚根的集合中 void Union(int x,int y) { int xx=find(x),yy=find(y); if(xx!=yy) node[xx].father=yy; } //删除节点时，只要把他的父节点置为一个无用的虚根 void deleteX(int x) { node[x].replace=extra; node[node[x].replace].father=extra; extra++; } int main() { int n,m,k=1,x,y,sum; char c[2]; while(scanf("%d%d",&n,&m),n||m){ init(n); for(int i=0;i<m;i++) { scanf("%s",c); if(c[0]=='M') { scanf("%d%d",&x,&y); Union(node[x].replace,node[y].replace); } else { scanf("%d",&x); deleteX(x); } } set<int>ans; // 计算集合的个数用set ans.clear(); for(int i=0;i<n;i++) { int temp=find(node[i].replace); ans.insert(temp); } printf("Case #%d: %d\n",k++,ans.size()); } return 0; }
