poj2492 A Bug's Life【并查集】

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!

Hint

Huge input,scanf is recommended.

假设虫子只和异性交流 现在告诉你交流的虫子号码 问这个假设正确不正确

现在发现并查集就是找节点和父节点的关系 rank数组就是x对fx的关系 有时候不考虑方向比如这一题

像这种题目更新的时候 fx的rank应该是 x对fx的关系加fy对y的关系加x对y的关系

输出之间要空行！

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;

const int maxn = 2005;
int t, n, m;
int ran[maxn], parent[maxn];

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

int fin(int x)
{
    if(x == parent[x]) return x;

    int t = parent[x];
    parent[x] = fin(parent[x]);
    ran[x] = (ran[x] + ran[t]) % 2;
    return parent[x];
}

void mer(int x, int y)
{
    int tx = fin(x);
    int ty = fin(y);

    if(tx != ty){
        parent[tx] = ty;
        ran[tx] = (ran[x] + 1 + ran[y]) % 2;
    }

}

int main()
{
    scanf("%d", &t);
    for(int i = 1; i <= t; i++){
        scanf("%d%d", &n, &m);
        init(n);
        bool flag = true;
        for(int j = 0; j < m; j++){
            int a, b;
            scanf("%d%d", &a, &b);
            if(fin(a) == fin(b)){
                if(ran[a] == ran[b])
                {
                    flag = false;
                }
            }
            else{
                mer(a, b);
            }
        }

        if(!flag){
            printf("Scenario #%d:\nSuspicious bugs found!\n\n", i);
        }
        else{
            printf("Scenario #%d:\nNo suspicious bugs found!\n\n", i);
        }
    }
    return 0;
}