hdoj-1289-A Bug's Life【种类并查集】
A Bug's Life
Time Limit: 15000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 11386 Accepted Submission(s): 3709
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.
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.
about the bugs' sexual behavior, or "Suspicious bugs found!" if Professor Hopper's assumption is definitely wrong.
2
3 3
1 2
2 3
1 3
4 2
1 2
3 4
Scenario #1:
Suspicious bugs found! Scenario #2:
No suspicious bugs found!HintHuge input,scanf is recommended.
pid=1558" target="_blank">
1558
pid=1811" target="_blank">
1811
pid=1325" target="_blank">
1325
#include<stdio.h>
int root[2010];
int rank[2010];
int find_set(int x){
if(root[x]==x) return x;
int t=root[x];
root[x]=find_set(t);
rank[x]=(rank[x]+rank[t])%2;
return root[x];
}
void unio(int a,int b){
int t,k;
t=root[a];
k=root[b];
root[t]=k;
rank[t]=(rank[b]+1-rank[a]);
return ;
}
int main(){
int t,cas=0;;
scanf("%d",&t);
while(t--){
printf("Scenario #%d:\n",++cas);
int n,m,i,ok=0,a,b;
scanf("%d%d",&n,&m);
for(i=0;i<=n;++i){
root[i]=i;
rank[i]=0;
}
for(i=0;i<m;++i){
scanf("%d%d",&a,&b);
if(!ok){
int pa,pb;
pa=find_set(a) ;
pb=find_set(b);
if(pa==pb&&(rank[a]==rank[b])){
ok=1;
}
else
unio(a,b);
}
}
if(ok)
printf("Suspicious bugs found!\n\n");
else
printf("No suspicious bugs found!\n\n"); }
return 0;
}
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