A Bug's Life

Time Limit: 15000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 11981 Accepted Submission(s): 3901

Problem Description

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

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.

并查集题目。

记录一下当前结点到根结点的距离。当再次找进行”并“操作时，判断属于同一根节点的两个节点到根节点的距离，同时为奇数或偶数，那么久存在矛盾。

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                      #define ull unsigned long long#define ll long long#define mod 90001#define INF 0x3f3f3f3f#define maxn 3000+10#define cle(a) memset(a,0,sizeof(a))const ull inf = 1LL << 61;const double eps=1e-5;using namespace std;bool cmp(int a,int b){ return a>b;}int fa[maxn];int d[maxn];int n,m,mark;void init(){ for(int i=1;i<=n+5;i++){ d[i]=0; fa[i]=i; } mark=0;}int findfa(int x){ if(x==fa[x])return x; else{ int root=findfa(fa[x]); d[x]+=d[fa[x]];//记录到根节点的距离 fa[x]=root; return fa[x]; }}void Union(int a,int b){ int x=findfa(a); int y=findfa(b); if(x==y){ if((d[a]&1)==(d[b]&1))mark=1; return ; } else{ fa[x]=y; d[x]=(1+d[b]-d[a]);//向量 }}int main(){ #ifndef ONLINE_JUDGE freopen("in.txt","r",stdin); #endif //freopen("out.txt","w",stdout); int T; cin>>T; int a,b; int cnt=1; while(T--){ scanf("%d%d",&n,&m); init(); for(int i=0;i

转载于:https://www.cnblogs.com/pk28/p/4732059.html

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