hdu 1829 並查集（食物鏈的弱化版）

hdu 1829 並查集（食物鏈的弱化版）

 

 

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!

HintHuge input,scanf is recommended. 

/**
hdu 1829  並查集（食物鏈的弱化版）
題目大意：給定n個昆蟲，和m種戀愛關系，判斷是否有同性戀
解題思路：並查集i-A表示i屬於性別A若i-A和j-B,屬於同一類，表示i為性別A和j為性別B是同時發生或者同時不發生。類似於poj1182
*/
#include 
#include 
#include 
#include 
using namespace std;
const int maxn=2005;
int par[maxn*2],_rank[maxn*2];

void init(int n)
{
    for(int i=0; i<=n; i++)
    {
        par[i]=i;
        _rank[i]=0;
    }
}

int find(int x)
{
    if(par[x]==x)
        return x;
    return par[x]=find(par[x]);
}

void unite(int x,int y)
{
    x=find(x);
    y=find(y);
    if(x==y)return;
    if(_rank[x]<_rank[y])
    {
        par[x]=y;
    }
    else
    {
        par[y]=x;
        if(_rank[x]==_rank[y])
            _rank[x]++;
    }
}

bool same(int x,int y)
{
    return find(x)==find(y);
}

int main()
{
    int T,tt=0,n,m;
    scanf(%d,&T);
    while(T--)
    {
        int flag=0;
        scanf(%d%d,&n,&m);
        init(n*2);
        while(m--)
        {
            int x,y;
            scanf(%d%d,&x,&y);
            if(same(x,y)||same(x+n,y+n))
            {
                flag=1;
            }
            else
            {
                unite(x,y+n);
                unite(x+n,y);
            }
        }
        printf(Scenario #%d:
,++tt);
        if(flag)
            puts(Suspicious bugs found!
);
        else
            puts(No suspicious bugs found!
);
    }
    return 0;
}