偏移向量并查集——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.
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.**

#include<cstdio>#include<iostream>using namespace std;int par[2005];int d[2005];   //偏移向量//const int same = 0, diff = 1;int flag;/*int find(int x){    int t;    if (x != par[x])//    {        t = find(par[x]);        d[x] = (d[x] + d[par[x]]) %2;        return t;    }    return par[x];}*/int find(int x){    int t;    if (x != par[x])    {        t = par[x];        par[x]= find(par[x]);        //路径压缩        d[x] = (d[x] + d[t]) % 2;    }    return par[x];}void init(int n){    for (int i = 1; i <= n; i++)    {        par[i] = i;        d[i] = 0;    }    flag = 1;}void unite(int a, int b){    int a_par = find(a);    int b_par = find(b);    if (a_par != b_par)    {        par[a_par] = b_par;        //d[a_par] = diff;  不可        d[a_par] = (d[b] + 1 - d[a]) % 2;    }    find(a);}int main(){    int T, cnt = 0;    cin >> T;    while (T--)    {        int n, m;        cin >> n >> m;        init(n);        for (int i = 0; i < m; i++)        {            int a, b;            cin >> a >> b;             if(flag)                unite(a, b);            if (d[a]==d[b])            {                flag = 0;                //break;       此处不应break              }        }            cout << "Scenario #" << ++cnt << ":" << endl;            if (flag)                cout << "No suspicious bugs found!" << endl;  //注意大小写            else                cout << "Suspicious bugs found!" << endl;            cout << endl;    }    return 0;}