A Bug's Life
Time Limit: 15000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8826 Accepted Submission(s): 2845
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.
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.
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!
/*
题目意思就是判断是否为同性
给定一系列数对，例如a和b，表示a和b不是同一种性别
检查一堆数据中是否有同性恋，
找出主要矛盾是如果1喜欢2,2喜欢3，而1又喜欢3，则1 3 同性矛盾。
找出出现类的的矛盾
rank[]用作记录某点跟根节点的距离，
rank表示当前点到父节点的距离，如果距离是奇数，那么这两个就是异性，否则是同性，
rank不是1就是0 也可以一认为与根节点关系
这里用它来记录某点跟根节点的关系，所以得到距离模2即可。
*/
#include<iostream>
#include<stdio.h>
using namespace std;
int pre[2001],rank[2001];
int find(int x)
{
int t;
if(x==pre[x])
return x;
t=find(pre[x]);// 找到根节点
rank[x]=(rank[x]+rank[pre[x]])%2;//更新该节点与要更新的节点关系
pre[x]=t;
return t;
}
int join(int x,int y)
{
int a=find(x);
int b=find(y);//二者根
if(a==b)
{
if(rank[x]==rank[y])//根相同 而且与根节点的性别的区别也一样，必然同性
return 1;
return 0;
}
pre[a]=b;//二者根不同 更新使根a,b有相同的根b
//则b不用变 更新a的rank 判断与根b的关系
rank[a]=(rank[x]+rank[y]+1)%2;//x与y异性 所以+1
//rank[x]表示x与a的关系 rank[y]表示y与b的关系
//例如：1+1 x与a异性 y与b异性 x与y异性 a与b异性 rank[a]应该为异性1
//例如：1+0 x与a异性 y与b同性 x与y异性 a与b同性 rank[a]应该为同性0
return 0;
}
int main(){
int t,n,m,c=1,i,flag,a,b;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&m);
memset(rank,0,sizeof(rank));
for(i=1;i<=n;i++)
pre[i]=i;
flag=0;
for(i=0;i<m;i++)
{
scanf("%d%d",&a,&b);
if(join(a,b))
flag=1;
}
printf("Scenario #%d:\n",c++);
if(flag)
printf("Suspicious bugs found!\n");
else
printf("No suspicious bugs found!\n");
printf("\n");
}
return 0;
}