hdu1829 a bug's life
来源:互联网 发布:html编写软件 编辑:程序博客网 时间:2024/06/08 17:58
A Bug's Life
Time Limit: 15000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7176 Accepted Submission(s): 2330
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.
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 | We have carefully selected several similar problems for you: 1558 1198 1325 1213 2473
1 //796MS 240K 1104 B G++ 2 /* 3 4 题意: 5 给出m对关系,每对都是异性,问有木有矛盾的 6 7 并查集: 8 开始自己写的集中思路都不对,其实就是要处理一个二分图, 9 如果该二分图存在矛盾时就为有矛盾,具体思路:10 opp保存对立单位,set保并查集中的元素;11 当opp[x]==0 && opp[y]==0时,证明x、y无对立关系,设置对立关系:opp[x]=y;opp[y]=x;12 当opp[x]==0 && opp[y]!=0时,设置x对立关系,并合并opp[y]与x:opp[x]=y;merge(x,opp[y]);13 当opp[x]!=0 && opp[y]==0时,同上:opp[y]=x;merge(y,opp[x]); 14 当opp[x]!=0 && opp[y]!=0时,15 如find(x)==find(b),证明存在矛盾16 否则将x、y与其对立关系的合并:merge(y,opp[x]);merge(x,opp[y]); 17 18 */19 #include<stdio.h>20 int set[2005];21 int find(int x)22 {23 if(x==set[x]) return x;24 return find(set[x]); 25 }26 void merge(int a,int b)27 {28 set[find(a)]=find(b);29 }30 int main(void)31 {32 int t,n,m;33 int x,y,k=1;34 scanf("%d",&t);35 while(t--)36 {37 int flag=0;38 int opp[2005]={0};39 scanf("%d%d",&n,&m);40 for(int i=1;i<=n;i++) set[i]=i;41 for(int i=0;i<m;i++){42 scanf("%d%d",&x,&y);43 if(!flag){44 if(opp[x]==0 && opp[y]==0){45 opp[x]=y;opp[y]=x;46 }else if(opp[x]==0){47 opp[x]=y;merge(x,opp[y]);48 }else if(opp[y]==0){49 opp[y]=x;merge(y,opp[x]);50 }else{51 if(find(x)==find(y)) flag=1;52 merge(y,opp[x]);merge(x,opp[y]);53 }54 }55 }56 printf("Scenario #%d:\n",k++);57 if(flag) printf("Suspicious bugs found!\n");58 else printf("No suspicious bugs found!\n");59 printf("\n");60 }61 return 0;62 }
阅读全文
0 0
- hdu1829 A Bug's Life
- hdu1829 A Bug's Life
- hdu1829 A Bug's Life
- hdu1829 A Bug's Life
- HDU1829- A Bug's Life
- hdu1829 a bug's life
- HDU1829 POJ2492 A Bug's Life
- HDU1829 A Bug's Life 并查集
- HDU1829(A Bug's Life )—并查集
- A Bug's Life hdu1829 并查集
- A Bug's Life(hdu1829种类并查集)
- hdu1829 A Bug's Life (并查集)
- hdu1829 A Bug's Life 并查集
- hdu1829 A Bug's Life(并查集)
- HDU1829 - A Bug's Life 分组并查集
- hdu1829 A Bug's Life --分组并查集
- HDU1829:A Bug's Life(并查集)
- HDU1829:A Bug's Life(种类并查集)
- oracleday07(三层过滤 自复制 分页查询)
- [1] 微信公众号开发
- 三分钟学会逻辑回归
- [Zabbix] Tomcat 监控
- 设计模式的五大原则
- hdu1829 a bug's life
- android和ios的数组逆序
- DOS下cd命令的使用说明
- poj1611The Suspects
- Error creating bean with name 'shiroFilter' defined in class path resource
- 键盘上所有特殊符号的英文读法 来自: 2010-07-02 01:08:53 ! 叹号 exclamation mark/bang ? 问号 question mark , 逗号 comm
- 字符串拼接方式性能比较
- js中获取时间new Date()详细介绍
- error: The following untracked working tree files would be overwritten by merge