[并查集]HDU 1829/POJ 2492/HOJ 2319 A Bug's Life

传送门：A Bug's Life

A Bug's Life

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

Total Submission(s): 6961    Accepted Submission(s): 2267


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

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!

HintHuge input,scanf is recommended.

 

Source

TUD Programming Contest 2005, Darmstadt, Germany

解题报告：

这个题属于并查集中求偏移量的问题，给你的2个人的关系，已知他们是异性，要求你判断其中是否可能存在同性恋。
代码如下：
#include<iostream>
#include<cstdio>
using namespace std;
int m,n,flag;
int father[20005],r[20005];
void makeset(int n){
for(int j=1;j<=n;j++){
father[j]=j;
r[j]=0;
}
}
int findset(int x){
int a=father[x];
if(father[x]==x)
return x;
father[x]=findset(father[x]);
r[x]=(r[x]+r[a])%2;
return father[x];
}
void Union(int a,int b){
if(flag)
return;
int x=findset(a);
int y=findset(b);
if(x==y&&r[a]==r[b]){
flag=1;
return;
}
father[x]=y;
r[x]=(r[a]+r[b]+1)%2;
}
int main(){
int t,s=0;
scanf("%d",&t);
while(t--){
scanf("%d%d",&n,&m);
flag=0;
makeset(n);
for(int i=0;i<m;i++){
int a,b;
scanf("%d%d",&a,&b);
Union(a,b);
}
printf("Scenario #%d:\n",++s);
flag?puts("Suspicious bugs found!\n"):puts("No suspicious bugs found!\n");
}
return 0;
}