hdu 1213 How Many Tables(简单的并查集)
2017-03-19 13:46
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[align=left]Problem Description[/align]
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.
[align=left]Input[/align]
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.
[align=left]Output[/align]
For each test case, just output how many tables Ignatius needs at least. Do NOT print any blanks.
[align=left]Sample Input[/align]
2
5 3
1 2
2 3
4 5
5 1
2 5
[align=left]Sample Output[/align]
2
4
[align=left]Author[/align]
Ignatius.L
[align=left]Source[/align]
杭电ACM省赛集训队选拔赛之热身赛
题意我没有读直接看样例猜的,一下子过
思路 : 就是没次输入a和b时,将a b合并 ,最后找每个点的根节点,用数组记录下来,去重即可
代码
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.
[align=left]Input[/align]
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.
[align=left]Output[/align]
For each test case, just output how many tables Ignatius needs at least. Do NOT print any blanks.
[align=left]Sample Input[/align]
2
5 3
1 2
2 3
4 5
5 1
2 5
[align=left]Sample Output[/align]
2
4
[align=left]Author[/align]
Ignatius.L
[align=left]Source[/align]
杭电ACM省赛集训队选拔赛之热身赛
题意我没有读直接看样例猜的,一下子过
思路 : 就是没次输入a和b时,将a b合并 ,最后找每个点的根节点,用数组记录下来,去重即可
代码
#include<stdio.h> #include<string.h> #include<stdlib.h> #include<ctype.h> #include<math.h> #include<stack> #include<queue> #include<map> #include<set> #include<vector> #include<string> #include<iostream> #include<algorithm> #include<utility> #include<iomanip> #include<time.h> typedef long long ll; const double Pi = acos(-1.0); const int N = 1e6+10, M = 1e3+20, mod = 1e9+7, inf = 2e9+10; const double e=2.718281828459 ; const double esp=1e-9; using namespace std; int n,m; int Father[M]; int s[M]; struct node { int a,b; } aa[M]; int Find(int x) { if(x!=Father[x]) return Father[x]=Find(Father[x]); else return Father[x]; } void Union(int x,int y) { int p1=Find(x); int p2=Find(y); if(p1!=p2) Father[p1]=p2; } int main() { int t; scanf("%d",&t); while(t--) { memset(s,0,sizeof(s)); scanf("%d%d",&n,&m); // int ns; for(int i=1; i<=n; i++) { Father[i]=i; } for(int i=0; i<m; i++ ) { scanf("%d%d",&aa[i].a,&aa[i].b); Union(aa[i].a,aa[i].b); } // int s[M]; for(int i=1;i<=n;i++) { s[i-1]=Find(i); } sort(s,s+n); int ns=unique(s,s+n)-s; printf("%d\n",ns); } return 0; }
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