hdu 1213 How Many Tables(并查集算法)
来源:互联网 发布:北京全国接单淘宝贷款 编辑:程序博客网 时间:2024/06/04 18:39
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1213
How Many Tables
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13787 Accepted Submission(s): 6760
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
2
5 3
1 2
2 3
4 5
5 1
2 5
Sample Output
2
4
题目大意:陌生的人不会坐在一起,也就是最短路的问题,两个人之间直接有路或者是间接有路即可,找到最少需要建路的条数(即本题的桌子数)。这个题目要判断能否找到,所以需要判断哦~~
详见代码。
1 #include <iostream> 2 #include <cstdio> 3 using namespace std; 4 int father[1010]; 5 6 void set(int n) 7 { 8 for (int i=1; i<=n; i++) 9 father[i]=i;10 }11 12 int find(int a)13 {14 if (father[a]==a)15 return a;16 return father[a]=find(father[a]);17 }18 19 void Union(int x,int y)20 {21 x=find(x);22 y=find(y);23 if (x!=y)24 father[x]=y;25 }26 27 int main ()28 {29 int t,k;30 while (cin>>t)31 {32 33 while (t--)34 {35 k=0;36 int n,m;37 cin>>n>>m;38 set(n);39 while (m--)40 {41 int s,d;42 cin>>s>>d;43 Union(s,d);44 }45 for (int i=1; i<=n; i++)46 {47 if (father[i]==i)48 k++;49 }50 printf ("%d\n",k);51 }52 }53 return 0;54 }
0 0
- hdu 1213 How Many Tables(并查集算法)
- HDU 1213 How Many Tables(并查集)
- hdu 1213 How Many Tables(并查集练习)
- [ACM] hdu 1213 How Many Tables(并查集)
- [ACM] hdu 1213 How Many Tables(并查集)
- hdu 1213 How Many Tables(并查集学习)
- HDU-#1213 How Many Tables (并查集)
- HDU - 1213 How Many Tables (简单并查集)
- HDU 1213 How Many Tables(并查集)
- HDU 1213 How Many Tables (并查集)
- HDU 1213【】How Many Tables(并查集)
- hdu 1213 How Many Tables(并查集)
- hdu 1213 How Many Tables (并查集)
- HDU 1213 How Many Tables(并查集)
- HDU-1213 How Many Tables(并查集)
- HDU 1213 How Many Tables(并查集)
- HDU 1213 How Many Tables(并查集)
- HDU 1213 How Many Tables(并查集)
- hdu 1875 畅通工程再续
- hdu 1230 火星A+B
- Android 项目(一):网络请求封装(一)
- hdu 1863 畅通工程 (并查集+最小生成树)
- KVO实现机制
- hdu 1213 How Many Tables(并查集算法)
- hdu 1232 畅通工程(并查集算法)
- hdu 1272 小希的迷宫(并查集+最小生成树+队列)
- hdu 1879 继续畅通工程 (并查集+最小生成树)
- 装饰者模式(decorator)
- hdu 1598 find the most comfortable road (并查集+枚举)
- UIWebView
- gVIM学习笔记1-基本命令
- hdu 3371(prim算法)