HDU 1829 A Bug's Life 经典带权并查集

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

23 31 22 31 34 21 23 4

Sample Output

Scenario #1:Suspicious bugs found!Scenario #2:No suspicious bugs found!

Hint

Huge input,scanf is recommended.

题意：有T组数据，有n个节点吗，m个关系。。

          1 2 为异性，2 3 为异性 按照逻辑1 3 应该为同性， 而 输入1 3 与逻辑上不符合，发现可疑。

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#include<cstdio>#include<algorithm>#include<cstring>using namespace std;const int M=2010;int p[M],r[M];int t;int n,m;void init(){    for(int i=0;i<=n;i++)    {        p[i]=i;        r[i]=0;    }}int finds(int x){    if(x==p[x])        return x;    int t=finds(p[x]);    r[x]=(r[x]+r[p[x]])%2;    p[x]=t;    return p[x];}bool unions(int a,int b){    int x=finds(a);    int y=finds(b);    if(x==y)    {        if(r[a]==r[b])            return 1;        else            return 0;    }    p[y]=x;    r[y]=(r[a]+r[b]+1)%2;    return 0;}int main(){    int a,b,k=0;    scanf("%d",&t);    while(t--)    {        scanf("%d%d",&n,&m);        init();        int ans=0;        for(int i=0;i<m;i++)        {            scanf("%d%d",&a,&b);            if(unions(a,b))                ans++;        }        printf("Scenario #%d:\n",++k);        if(ans>0)        printf("Suspicious bugs found!\n");        else            printf("No suspicious bugs found!\n");            printf("\n");    }    return 0;}