D - How Many Tables (并查集)(水题)
Today is Ignatius' birthday. He invites a lot of friends. Now it's dinner time. Ignatius wants to know how many tables he needs at least. You have to notice that not all the friends know each other, and all the friends do not want
to stay with strangers.
One important rule for this problem is that if I tell you A knows B, and B knows C, that means A, B, C know each other, so they can stay in one table.
For example: If I tell you A knows B, B knows C, and D knows E, so A, B, C can stay in one table, and D, E have to stay in the other one. So Ignatius needs 2 tables at least.
One important rule for this problem is that if I tell you A knows B, and B knows C, that means A, B, C know each other, so they can stay in one table.
For example: If I tell you A knows B, B knows C, and D knows E, so A, B, C can stay in one table, and D, E have to stay in the other one. So Ignatius needs 2 tables at least.
2 5 3 1 2 2 3 4 5 5 1 2 5
2 4
这个为什么TEL
Select Code
#include <stdio.h>
#include <string.h>
#define Maxn 1005
int par[Maxn],count=0;
void init(int n)
{
for(int i=0; i<n; i++)
par[i]=i;
}
int find(int x)
{
int r=x;
while(par[r]!=r)
{
r=par[r];
}
int i=x,j;
while(i!=r)
{
j=par[r];
par[r]=i;
i=j;
}
return r;
}
void join(int x, int y)
{
int fx=find(x), fy=find(y);
if(fx!=fy)
{
par[fx]=fy;
count++;
}
}
int main()
{
int i,j,a,b,n;
scanf("%d",&n);
while(n--)
{
count=0;
scanf("%d%d",&a,&b);
init(a);
while(b--)
{
scanf("%d%d",&i,&j);
join(i,j);
}
printf("%d\n",a-count);
}
return 0;
}
AC代码
#include <iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
using namespace std;
int pre[1100];
int findset(int v)
{
int t1,t2=v;
while(v!=pre[v]) v=pre[v];
while(t2!=pre[t2])
{
t1=pre[t2];
pre[t2]=v;
t2=t1;
}
return v;
}
void unions(int x,int y)
{
int t1=findset(x);
int t2=findset(y);
if(t1!=t2) pre[t1]=t2;
}
int main()
{
int T,n,m;
cin>>T;
while(T--)
{
cin>>n>>m;
for(int i=1;i<=n;i++) pre[i]=i;
for(int i=0;i<m;i++)
{
int u,v;
scanf("%d%d",&u,&v);
unions(u,v);
}
int ans=0;
for(int i=1;i<=n;i++)
if(pre[i]==i) ans++;
cout<<ans<<endl;
}
}
永远渴望,大智若愚(stay hungry, stay foolish)