hdu 1829 A Bug's Life（分组并查集）

A Bug's Life

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

Total Submission(s): 11242 Accepted Submission(s): 3664



[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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.

[align=left]Output[/align]
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.

[align=left]Sample Input[/align]

2
3 3
1 2
2 3
1 3
4 2
1 2
3 4

[align=left]Sample Output[/align]

Scenario #1:
Suspicious bugs found!

Scenario #2:
No suspicious bugs found!

HintHuge input,scanf is recommended.
题意： 给出昆虫的数量n和m个关系，输入a b即为a喜欢b，假设虫子只喜欢异性，问根据这一些列行为是否能发现同性恋的虫子
思路: 并查集的使用和理解，，思路好难想，下边是理解大神代码后的注释
2015,7,22

#include<stdio.h>
#include<string.h>
struct node{
int from;//父节点 即为自己的性别
int conflict;//用来存对象的性别
}f[2100];
int find(int x)
{
if(f[x].from!=x)
x=find(f[x].from);
return x;
}
void merge(int a,int b)
{
int x=find(a);
int y=find(b);
if(x!=y)
f[x].from=y;
}
int main()
{
int t,n,m,flag,Case=1,i,a,b;
scanf("%d",&t);
while(t--){
flag=0;
memset(f,0,sizeof(f));
scanf("%d%d",&n,&m);
for(i=0;i<=n;i++)
{
f[i].from=i;//父节点都初始化为自己
f[i].conflict=0;//开始初始化都没有对象
}
for(i=0;i<m;i++)
{
scanf("%d%d",&a,&b);//给出两个相互为对象
int x=find(a);
int y=find(b);
if(x==y) flag=1;//他们的父节点一样，那就是他们在一个集合里，就属于同性恋
if(f[a].conflict!=0)//如果a有对象
merge(b,f[a].conflict);//就把b拉到a对象的那个性别集合中
if(f[b].conflict!=0)//如果b有对象
merge(a,f[b].conflict);//把a拉到b对象的那个性别集合中
f[a].conflict=b;//把a的对象设置为b
f[b].conflict=a;//把b的对象设置为a
}
if(flag==1)
printf("Scenario #%d:\nSuspicious bugs found!\n",Case++);
else printf("Scenario #%d:\nNo suspicious bugs found!\n",Case++);
printf("\n");
}
return 0;
}
//在贴个比较好懂的代码，思想差不多。
#include<stdio.h>
int mark[2020], a[2020];
int find(int x)
{
if(x != mark[x])
{
mark[x] = find(mark[x]);
}
return mark[x];
}
void merge(int x, int y)
{
x = find(x);
y = find(y);
if(x != y)
{
mark[x] = y;
}
}
int main()
{
int i, j, k, m, n, T, x, y, ok, cas = 1;
scanf("%d", &T);
while(T--)
{
scanf("%d%d", &n, &m);
ok = 0;
for(i = 0; i <= n+1; i++)
{
mark[i] = i;
a[i] = -1;
}
for(i = 0; i < m; i++)
{
scanf("%d%d", &x, &y);
if(ok)
continue;
if(a[x] != -1)//x有对象
{
if(a[y] != -1)// y有对象
{
if(find(x) == find(y))//若x和y在一个性别集合里
ok = 1;      //同性。。
merge(x,a[y]);//否则令彼此的对象都归为和彼此一样的性别集合内
merge(y,a[x]);
}
else//y没对象
{
a[y] = x;//设定y的对象
merge(y,a[x]);//令y和x的对象在一个性别集合里
}
}
else//x没对象
{
if(a[y] != -1)//y有对象
{
a[x] = y;//设定x的对象
merge(x, a[y]);//令x和y的对象在一个性别集合里
}
else
{
a[x] = y;//彼此设定对象
a[y] = x;
}
}
}
if(ok)
printf("Scenario #%d:\nSuspicious bugs found!\n", cas++);
else
printf("Scenario #%d:\nNo suspicious bugs found!\n", cas++);
printf("\n");
}
return 0;
}