（专题赛）A Bug's Life

最新推荐文章于 2022-04-14 19:45:52 发布

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最新推荐文章于 2022-04-14 19:45:52 发布

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原文链接：http://www.cnblogs.com/awsent/p/4266973.html

本文介绍了一个使用并查集算法解决生物学实验中虫性判断问题的方法。该问题来源于一个假设，即如果三只虫子两两之间的互动表明它们不可能全部为异性，则可能存在同性恋现象。通过构建并查集，可以有效地判断这一假设是否成立。

并查集

#转题意：Professor Hopper专门研究bug的生活习性，他表示若两只bugs的生活习性差别很大，则说明他们可能为不同的性别，但如果出现三只bugs的习性两两差别很大，则有可能出现同性恋的bug了。现在有n只bugs，和生活习性差别很大的m对bugs的编号，问这些bugs中，有没有可能出现同性恋者。题目中给出的数对比如（1 2 ，2 3 ，1 3）是表示交配关系的，而且交配的都默认为理解为异性，这里如果有矛盾的情况，就表明有同性恋存在。比如1和2是异性，2和3为异性，那就表明1和3是同性，但是给出的数对是1和3异性表明有矛盾存在，则有同性恋存在，则输出Suspicious bugs found。

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

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!

  1 #include<stdio.h>
  2 
  3 
  4 
  5 int pre[1000],rank[1000];
  6 
  7 
  8 
  9 int find(int x)
 10 
 11 {
 12 
 13 if(x!=pre[x])
 14 
 15 pre[x]=find(pre[x]);
 16 
 17 return pre[x];
 18 
 19 }
 20 
 21 
 22 
 23 void join(int i,int j)
 24 
 25 {
 26 
 27 int x=find(i);
 28 
 29 int y=find(j);
 30 
 31 if(x==y)
 32 
 33 return;
 34 
 35 else
 36 
 37 pre[y]=x;
 38 
 39 }
 40 
 41 
 42 
 43 int main()
 44 
 45 {
 46 
 47 int t,i,m,n,x,y,ti=1;
 48 
 49 scanf("%d",&t);
 50 
 51 while(t--)
 52 
 53 {
 54 
 55 scanf("%d",&n);
 56 
 57 for(i=1;i<=n;i++)
 58 
 59 {
 60 
 61 pre[i]=i;
 62 
 63 rank[i]=0;
 64 
 65 }
 66 
 67 int flag=0;
 68 
 69 scanf("%d",&m);
 70 
 71 while(m--)
 72 
 73 {
 74 
 75 scanf("%d %d",&x,&y);
 76 
 77 if(flag)
 78 
 79 continue;
 80 
 81 if(find(x)==find(y))
 82 
 83 {//若x，y在同一个集合，则证明同性恋存在
 84 
 85 flag=1;
 86 
 87 }
 88 
 89 if(rank[x]==0&&rank[y]==0)
 90 
 91 {
 92 
 93 rank[x]=y; //表明y是x的异性
 94 
 95 rank[y]=x;
 96 
 97 }
 98 
 99 else if(rank[x]==0)
100 
101 {
102 
103 rank[x]=y;
104 
105 join(x,rank[y]);
106 
107 }
108 
109 else if(rank[y]==0)
110 
111 {
112 
113 rank[y]=x;
114 
115 join(y,rank[x]);
116 
117 }
118 
119 else
120 
121 {
122 
123 join(x,rank[y]);
124 
125 join(y,rank[x]);
126 
127 }
128 
129 }
130 
131 printf("Scenario #%d:\n",ti++);
132 
133 if(flag)
134 
135 puts("Suspicious bugs found!");
136 
137 else
138 
139 puts("No suspicious bugs found!");
140 
141 printf("\n");
142 
143 }
144 
145 return 0;
146 
147 }