A Bug's Life POJ

A Bug's Life

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.

 

题目：教授假设这种虫子有两种性别，且只有异性的虫子之间才有交流，每个虫子从1开始编号。接下来给出每对虫子之间的交流，问是不是只有异性之间才有交流。

思路：带权并查集板子题（不清楚带权并查集的点这里------>传送门）。同性权值为0，异性权值为1。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <vector>
#include <cmath>

using namespace std;

#define ll long long
#define pb push_back
#define fi first
#define se second

const int N = 2000 << 1;
struct node
{
    int rt, v;
}fa[N];

int Find(int x)
{
    if(fa[x].rt == x) return x;
    else{
        int tmp = fa[x].rt;
        fa[x].rt = Find(tmp);
        fa[x].v = (fa[x].v + fa[tmp].v) % 2;
        return fa[x].rt;
    }
}

bool Union(int x, int y)
{
    int fax = Find(x);
    int fay = Find(y);
    if(fax != fay){
        fa[fay].rt = fax;
        fa[fay].v = (fa[x].v + 1 - fa[y].v) % 2;
        return true;
    }else{
        if(0 == (fa[x].v + 1 - fa[y].v) % 2) return true;
        else return false;
    }
}

void solve()
{   
    int T;
    scanf("%d", &T);
    for(int _case = 1; _case <= T; ++_case){
        int n, m;
        scanf("%d%d", &n, &m);
        for(int i = 0; i <= n; ++i){
            fa[i].rt = i;
            fa[i].v = 0;
        }
        vector<pair<int ,int > > info;
        int x, y;
        for(int i = 1; i <= m; ++i){
            scanf("%d%d", &x, &y);
            info.pb({x, y});
        }

        int error = 0;
        for(int i = 0; i < m; ++i){
            x = info[i].fi;
            y = info[i].se;
            if(!Union(x, y)){
                error = 1;
                break;
            }
        }

        printf("Scenario #%d:
", _case);
        printf("%s

", error == 1 ? "Suspicious bugs found!" : "No suspicious bugs found!");
    }
}

int main()
{

    solve();

    return 0;
}