HDOJ 1829 A Bug's Life 并查集

并查集的加强训练

题目如下：

Problem Description

Background
Professor Hopper is researching the sexual behavior of a rare species of bugs. He assumes that they feature two different genders and that they only interact with bugs of the opposite gender. In his experiment, individual bugs and their interactions were easy to identify, because numbers were printed on their backs.

Problem
Given a list of bug interactions, decide whether the experiment supports his assumption of two genders with no homosexual bugs or if it contains some bug interactions that falsify it.

Input

The first line of the input contains the number of scenarios. Each scenario starts with one line giving the number of bugs (at least one, and up to 2000) and the number of interactions (up to 1000000) separated by a single space. In the following lines, each interaction is given in the form of two distinct bug numbers separated by a single space. Bugs are numbered consecutively starting from one.

Output

The output for every scenario is a line containing "Scenario #i:", where i is the number of the scenario starting at 1, followed by one line saying either "No suspicious bugs found!" if the experiment is consistent with his assumption about the bugs' sexual behavior, or "Suspicious bugs found!" if Professor Hopper's assumption is definitely wrong.

Sample Input

2 3 3 1 2 2 3 1 3 4 2 1 2 3 4

Sample Output

Scenario #1: Suspicious bugs found! Scenario #2: No suspicious bugs found!

Hint

Huge input,scanf is recommended.

Source

TUD Programming Contest 2005, Darmstadt, Germany

Recommend

linle

若 A->B 而且 A以前出现过，那么 A.opper与B一定是同性 即相同集合中

另注意，题目要求，输入数据完了以后才输出结果

/* 构造数组opp，初始化为本身编号，若A与B有关，首先进行find(A),find(B)的操作，若根相同，则说明A与B属于同一集合，即同性恋， 否则执行下面的操作，如果A的opp等于本身，说明A的opp未初始化，使之等于B，否则将A的opp与B进行Unition操作，类似的如果B的opp等于本身， 说明B的opp未初始化，使之等于A，否则将B的opp与A进行Unition操作 */ #include <iostream> #define N 2100 using namespace std; struct node { int father; int opper; }hash[N]; void init() // 初始化 { int i; for (i=0;i<N;i++) { hash[i].father=i; hash[i].opper=i; } } int Find(int x) //root 返回根节点 { int root=x,temp; while (root!=hash[root].father) { root=hash[root].father; } while (root!=hash[x].father) // 压缩路径 { temp=hash[x].father; hash[x].father=root; x=temp; } return root; } void Unition(int x,int y) // 合并 { int fx=Find(x); int fy=Find(y); hash[fy].father=fx; } /* 注释内是错误版本的主函数 int main() { int T,i; scanf("%d",&T); for(i=1;i<=T;i++) { int n,m,j,x,y,a,b;; scanf("%d%d",&n,&m); init(); bool flag=0; // 标记有没有发现同性恋 for (j=0;j<m;j++) { scanf("%d%d",&a,&b); x=Find(a); y=Find(b); if (x==y) // 存在同性恋 { flag=1; cout<<"Scenario #"<<i<<":"<<endl; cout<<"Suspicious bugs found!"<<endl; break; // 错误的地方， 虽然输入还没结束便可以确定有没有同性恋存在， // 但是按题目要求必须要等输入完成以后才能进行输出，否则后面输入的数据会紊乱 } if (hash[a].opper!=a) { Unition(y,hash[a].opper); } else { hash[a].opper=y; } if (hash[b].opper!=b) { Unition(x,hash[b].opper); } else { hash[b].opper=x; } } if(flag==0) { cout<<"Scenario #"<<i<<":"<<endl; cout<<"No suspicious bugs found!"<<endl; } cout<<endl; } } */ int main() { int T,i; scanf("%d",&T); for(i=1;i<=T;i++) { int n,m,j,x,y,a,b;; scanf("%d%d",&n,&m); init(); bool flag=0; // 标记有没有发现同性恋 for (j=0;j<m;j++) { scanf("%d%d",&a,&b); x=Find(a); y=Find(b); if(flag==1) { continue; } if (x==y) // 存在同性恋 { flag=1; } if (hash[a].opper!=a) { Unition(y,hash[a].opper); } else { hash[a].opper=y; } if (hash[b].opper!=b) { Unition(x,hash[b].opper); } else { hash[b].opper=x; } } if(flag==1) { cout<<"Scenario #"<<i<<":"<<endl; cout<<"Suspicious bugs found!"<<endl; } else { cout<<"Scenario #"<<i<<":"<<endl; cout<<"No suspicious bugs found!"<<endl; } cout<<endl; } }