How Many Tables
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 14 Accepted Submission(s) : 12
Font: Times New Roman | Verdana | Georgia
Font Size: ← →
Problem Description
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.
Input
The input starts with an integer T(1<=T<=25) which indicate the number of test cases. Then T test cases follow. Each test case starts with two integers N and M(1<=N,M<=1000). N indicates the number of friends,
the friends are marked from 1 to N. Then M lines follow. Each line consists of two integers A and B(A!=B), that means friend A and friend B know each other. There will be a blank line between two cases.
Output
For each test case, just output how many tables Ignatius needs at least. Do NOT print any blanks.
Sample Input
2 5 3 1 2 2 3 4 5 5 1 2 5
Sample Output
2 4
题目大意
有人请客聚餐，认识的人可以坐一桌，朋友的朋友可以坐一桌，如果这个桌子上一个人都不认识那么就得另坐一桌。
解题思路
这是考察了并查集的知识。
代码
#include<stdio.h>
int ren[1100];
struct relation
{
int sta,end;
}res[1100];
int relation(int a)
{
if(ren[a]==a)
return a;
else
return ren[a]=relation(ren[a]);
//retrn relation(ren[a])
//加上ren[a]可以优化路径
}
int main()
{
int t;
int n,m;
int sum;
int i,j,k,l;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&m);
sum=n;
for(i=1;i<=n;i++)
ren[i]=i;//初定义每个只跟自己有关系
for(i=0;i<m;i++)
scanf("%d%d",&res[i].sta,&res[i].end);
for(i=0;i<m;i++)
{
j=relation(res[i].sta);
k=relation(res[i].end);
//判断最初是不是一样
if(j!=k)
{
sum--;
ren[j]=k;
}
}
printf("%d\n",sum);
}
return 0;
}