并查集--poj 2492

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

Sample Output

Scenario #1:

Suspicious bugs found!

Scenario #2:

No suspicious bugs found!

Hint

Huge input,scanf is recommended.

题目大意：有N个虫子，M个关系，判断这些虫子的关系（即性取向）是否正确

　　　　　　按照样例输出。

解题思路：很明显用并查集比较好处理

　　　　　处理时候有一个就是保存深度的数组ra[Maxn]，0表示虫子是同性，1表示虫子是异性

下面就闲话少叙，代码走起

/*之前一直没搞懂为什么ra这个数组要取余运算原来是用0

    表示同类，1表示不同类，所以下面的21还有30行会取模*/

#include<iostream>

#include<cstdio>

using namespace std;

const int Maxn=2005;

int pa[Maxn],ra[Maxn];//用来存储父节点还有关系

void init(int n)

{

    for(int i=1;i<=n;i++)

    {

        pa[i]=i;

        ra[i]=0;

    }

}

int find_set(int x)

{

    int t=pa[x];

    if(pa[x]==x) return x;

    pa[x]=find_set(t);

    ra[x]=(ra[x]+ra[t])%2;

    return pa[x];

}

void union_set(int x,int y)

{

    int fx=find_set(x);

    int fy=find_set(y);

    if(fx==fy)    return;

    pa[fx]=fy;

    ra[fx]=(ra[x]+1+ra[y])%2;

}

int main()

{

    int T;

    int num,ncase;

    int x,y;

    scanf("%d",&T);

//    {

        for(int i=1;i<=T;i++)

        {

            bool flag=true;

            scanf("%d%d",&num,&ncase);

            init(num);

            while(ncase--)

            {

                scanf("%d%d",&x,&y);

                if(find_set(x)==find_set(y))

                {

                    if(ra[x]==ra[y])

                    {

                        flag=false;

                        continue;

                    }

                }

                else

                {

                    union_set(x,y);

                }

            }

        printf("Scenario #%d:\n", i);

        if(flag) printf("No suspicious bugs found!\n");

        else printf("Suspicious bugs found!\n");

        printf("\n");

        }

//    }

    return 0;

}

巴特西

并查集--poj 2492

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