zoukankan      html  css  js  c++  java
  • A Bug's Life 并查集,很好的应用,假定同性的放入一个集合中,矛盾时输出结果;

    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!
    ***************************************************************************************************************************
    ***************************************************************************************************************************
     1 #include<iostream>
     2 #include<string>
     3 #include<cstring>
     4 #include<cstdio>
     5 #include<cmath>
     6 using namespace std;
     7 int fa[2005];
     8 int com[2005];
     9 int i,j,k,n,m,cas;
    10 void init()
    11 {
    12     for(i=1;i<=2005;i++)
    13      fa[i]=i;
    14     memset(com,-1,sizeof(com));
    15 }
    16 int find(int x)
    17  {
    18      if(x==fa[x])
    19        return fa[x];
    20      return fa[x]=find(fa[x]);
    21  }
    22 void Unon(int x,int y)
    23  {
    24     int xt=find(x);
    25     int yt=find(y);
    26      if(xt!=yt)
    27       fa[xt]=yt;
    28  }
    29 int main()
    30 {
    31     scanf("%d",&cas);
    32     k=0;
    33     while(cas--)
    34     {
    35         ++k;
    36         init();
    37         int st,en;
    38         scanf("%d%d",&n,&m);
    39         int found=0;
    40         for(i=1;i<=m;i++)
    41         {
    42             scanf("%d%d",&st,&en);
    43             if(com[st]!=-1)
    44             {
    45                 if(com[en]!=-1)
    46                 {
    47                     if(find(st)==find(en))
    48                     {
    49                         found=1;
    50 
    51                     }
    52                     Unon(en,com[st]);
    53                     Unon(st,com[en]);
    54                      continue;
    55                 }
    56                 Unon(com[st],en);
    57                 com[en]=st;
    58             }
    59             else
    60             {
    61                 if(com[en]!=-1)
    62                 {
    63                     Unon(com[en],st);
    64                     com[st]=en;
    65                 }
    66                 else
    67                 {
    68                     com[st]=en;
    69                     com[en]=st;
    70                 }
    71             }
    72         }
    73         printf("Scenario #%d:
    ",k);
    74         if(found==1)
    75           printf("Suspicious bugs found!
    ");
    76         else
    77           printf("No suspicious bugs found!
    ");
    78         printf("
    ");
    79 
    80     }
    81     return 0;
    82 
    83 }
    View Code
     
  • 相关阅读:
    12-五子棋游戏:享元模式
    11-制作糖醋排骨:外观模式
    10-蒸馒头:装饰者模式
    09-公司层级结构:组合模式
    08-开关与电灯:桥接模式
    07-电源转换:适配器模式
    将博客搬至CSDN
    iview和element中日期选择器快捷选项的定制控件
    详解AJAX工作原理以及实例讲解(通俗易懂)
    最全肌肉锻炼动图
  • 原文地址:https://www.cnblogs.com/sdau--codeants/p/3375114.html
Copyright © 2011-2022 走看看