A Bug's Life POJ - 2492
2017-05-24 15:54
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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
Sample Output
Hint
Huge input,scanf is recommended.
思路:这题比较好理解,用x+n来表示能与x交配的所有编号的虫子,如果发现两个互相交配的虫子AB同时可以交配同一个虫子C,那证明有homo虫。
#include <stdio.h>
int parent[2005+2005];
int find(int x)
{
return (x==parent[x])?x:parent[x]=find(parent[x]);
}
void init(int a)
{
for(int i=1;i<=2*a;i++)
parent[i]=i;
}
int same(int x,int y)
{
return find(x)==find(y);
}
void unite(int x,int y)
{
x=find(x);
y=find(y);
if(x==y) return ;
else parent[y]=x;
}
int main(void)
{
int t;
scanf("%d",&t);
for(int i=1;i<=t;i++)
{
int flag=0;
int amount,pair;
scanf("%d%d",&amount,&pair);
init(amount);
for(int i=1;i<=pair;i++)
{
int num1,num2;
scanf("%d%d",&num1,&num2);
if(same(num1,num2)==1)
flag=1;
else
unite(num1+amount,num2),unite(num1,num2+amount); // 关键
}
if(flag)
printf("Scenario #%d:\nSuspicious bugs found!\n\n",i);
else
printf("Scenario #%d:\nNo suspicious bugs found!\n\n",i);
}
}
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.
思路:这题比较好理解,用x+n来表示能与x交配的所有编号的虫子,如果发现两个互相交配的虫子AB同时可以交配同一个虫子C,那证明有homo虫。
#include <stdio.h>
int parent[2005+2005];
int find(int x)
{
return (x==parent[x])?x:parent[x]=find(parent[x]);
}
void init(int a)
{
for(int i=1;i<=2*a;i++)
parent[i]=i;
}
int same(int x,int y)
{
return find(x)==find(y);
}
void unite(int x,int y)
{
x=find(x);
y=find(y);
if(x==y) return ;
else parent[y]=x;
}
int main(void)
{
int t;
scanf("%d",&t);
for(int i=1;i<=t;i++)
{
int flag=0;
int amount,pair;
scanf("%d%d",&amount,&pair);
init(amount);
for(int i=1;i<=pair;i++)
{
int num1,num2;
scanf("%d%d",&num1,&num2);
if(same(num1,num2)==1)
flag=1;
else
unite(num1+amount,num2),unite(num1,num2+amount); // 关键
}
if(flag)
printf("Scenario #%d:\nSuspicious bugs found!\n\n",i);
else
printf("Scenario #%d:\nNo suspicious bugs found!\n\n",i);
}
}
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