HDU---1829 A Bug's Life【并查集】

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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

23 31 22 31 34 21 23 4

Sample Output

Scenario #1:Suspicious bugs found!Scenario #2:No suspicious bugs found!

Source:点击打开链接

题意：

    给定一系列数对，例如a和b，表示a和b不是同一种性别，然后不断的给出这样的数对，问有没有性别不对的情况。
例如给定：
1    2
3    4
1    3
那这里就是说1和2不是同种性别，3和4也不是同种性别，1和3不是同种性别，那这样就说明1和3是同一种性别，2和4是同一种性别，所以没有任何歧义，这时候输出No suspicious bugs found!。
但是例如：
1 2
2 3
1 3
1和2不同性别，2和3不同性别，那么1和3同一性别的，但是第三组数对又表明1和3不同性别，所以这里就出现了3或者1的性别出现了歧义，也就是说条件矛盾，这时候输出Suspicious bugs found!
代码：
#include <iostream>#define maxn 2005using namespace std;int par[maxn];int relation[maxn];int find(int x);void unite(int x,int y);int main(){    int t,n,m,a,b;    int cas=1;    cin>>t;    while(t--)    {        bool flag=true;        cin>>n>>m;        for(int i=1; i<=n; i++)        {            par[i]=i;            relation[i]=1;//将根节点标记为1，之后的层层树枝为0,1，0，1，数字相同代表性别相同        }        for(int j=1; j<=m; j++)        {            cin>>a>>b;            if(find(a)==find(b)&&relation[a]==relation[b])//如果a与b属于同一组且数字相同，那就有bug                flag=false;            else                unite(a,b);        }            if(flag)            {                cout<<"Scenario #"<<cas++<<":"<<endl;                cout<<"No suspicious bugs found!"<<endl<<endl;            }            else            {                cout<<"Scenario #"<<cas++<<":"<<endl;                cout<<"Suspicious bugs found!"<<endl<<endl;            }    }}int find(int x){    if(par[x]==x)        return x;    int t=par[x];    par[x]=find(par[x]);    relation[x]=(relation[x]+relation[t]+1)%2;//压缩路径后重新计算距离    return par[x];}void unite(int x,int y){    int a=find(x);    int b=find(y);    par[a]=b;    relation[a]=(relation[y]-relation[x])%2;//连接后计算距离}<span class="Apple-style-span" style="font-weight: bold;"></span>

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