HDU-1213-How Many Tables
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How Many Tables
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 32673 Accepted Submission(s): 16300
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.
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
25 31 22 34 55 12 5
Sample Output
24
思路:直接并查集,集合的数量就是桌子的数量,所以只要找最后有多少个根节点就行了
#include<stdio.h>#include<string.h>using namespace std;int n,par[1010];bool flag[1010];//记录结点是否为根节点,若为1是根节点void init()//初始化{for(int i=1;i<=n;i++)par[i]=i;}int find(int x)//找根节点{return x==par[x]?x:par[x]=find(par[x]);}void Union(int a,int b)//把朋友放到一个集合(树)里{int fa=find(a),fb=find(b);if(fa!=fb)par[fa]=fb;}int main(){int T,g,i,a,b;scanf("%d",&T);while(T--){int ans=0;scanf("%d%d",&n,&g);init();for(i=1;i<=g;i++){scanf("%d%d",&a,&b);Union(a,b); }memset(flag,0,sizeof(flag));for(i=1;i<=n;i++)flag[find(i)]=1;//find(i)找到i的根节点,把根节点标记为1,非根节点标记为0for(i=1;i<=n;i++)if(flag[i])ans++;printf("%d\n",ans);}return 0;}
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