zoukankan      html  css  js  c++  java
  • *HDU1829 并查集

    A Bug's Life

    Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 14121    Accepted Submission(s): 4603


    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!
     
     
    题意:
    有若干bugs,一对一对的给出,判断其中是否有同性恋,如果1与2,2与3,1又与3,就说明1和3是gay.
    代码:
     1 //将相同性别的bugs放进同一并查集里,这样只要输入的两个bug有相同的根就是gay.
     2 //路径压缩的while 的find比递归的find稍快一些。
     3 #include<iostream>
     4 #include<cstdio>
     5 #include<cstring>
     6 using namespace std;
     7 int t,n,m;
     8 int fat[2005],sex[2005];
     9 int find(int x)
    10 {
    11     int rt=x;
    12     while(fat[rt]!=rt)
    13     rt=fat[rt];
    14     int i=x,j;
    15     while(i!=rt)
    16     {
    17         j=fat[i];
    18         fat[i]=rt;
    19         i=j;
    20     }
    21     return rt;
    22 }
    23 /*int find(int x)
    24 {
    25     if(fat[x]!=x)
    26     fat[x]=find(fat[x]);
    27     return fat[x];
    28 }*/
    29 void connect(int x,int y)
    30 {
    31     int xx=find(x),yy=find(y);
    32     if(xx!=yy)
    33     fat[xx]=yy;
    34 }
    35 int main()
    36 {
    37     int a,b;
    38     scanf("%d",&t);
    39     for(int k=1;k<=t;k++)
    40     {
    41         scanf("%d%d",&n,&m);
    42         for(int i=0;i<=n;i++)
    43         {
    44             fat[i]=i;
    45             sex[i]=0;
    46         }
    47         bool flag=0;
    48         for(int i=1;i<=m;i++)
    49         {
    50             scanf("%d%d",&a,&b);
    51             if(flag) continue;
    52             if(find(a)==find(b))
    53             {
    54                 flag=1;
    55                 continue;
    56             }
    57             if(sex[a]==0) sex[a]=b;
    58             else connect(sex[a],b);
    59             if(sex[b]==0) sex[b]=a;
    60             else connect(sex[b],a);
    61         }
    62         printf("Scenario #%d:
    ",k);
    63         if(flag) printf("Suspicious bugs found!
    ");
    64         else printf("No suspicious bugs found!
    ");
    65         printf("
    ");
    66     }
    67     return 0;
    68 }
  • 相关阅读:
    竞赛中遇到的英文单词集锦
    #复习 搜索与图论:排列数字、走迷宫~ 20.8.20起
    #容斥原理 20.9.10
    #sort :快速排序、第k个数 20.09.12
    #STL #unordered_map : C++11 unordered_map详细介绍
    Java 面试必备(字符串专题)
    JAVA面试必备题(垃圾回收专题)
    Collection框架的结构(面试重点)
    CSS 层级样式表
    window搭建mongodb副本集
  • 原文地址:https://www.cnblogs.com/--ZHIYUAN/p/6013686.html
Copyright © 2011-2022 走看看