HDU 1213 How Many Tables 简单并查集
Problem DescriptionToday 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.
InputThe 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.
OutputFor each test case, just output how many tables Ignatius needs at least. Do NOT print any blanks.
Sample Input25 31 22 34 55 12 5
Sample Output24
AuthorIgnatius.L
Source杭电ACM省赛集训队选拔赛之热身赛
RecommendEddy
查了再并...就是这么easy...
#include <set>#include <list>#include <cmath>#include <ctime>#include <deque>#include <queue>#include <stack>#include <cctype>#include <cstdio>#include <string>#include <vector>#include <cassert>#include <cstdlib>#include <cstring>#include <sstream>#include <iostream>#include <algorithm>using namespace std;int father[1001];int sum;int find(int x){if(x!=father[x])father[x] = find(father[x]);//回溯,所有节点都变为祖宗节点的子节点...即树的深度=2return father[x];}void Union(int a,int b){int x = find(a);int y = find(b);if(x==y)return;sum--;father[y]=x;}int main(){//freopen("in.txt","r",stdin);int x;cin>>x;while(x--){memset(father,0,sizeof(father));int n,m,i,j;cin>>n>>m;for(i=1;i<=1001;i++)father[i]=i;sum=n;while(m--){int a,b;cin>>a>>b;Union(a,b);}cout<<sum<<endl;} return 0;}
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.
InputThe 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.
OutputFor 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
AuthorIgnatius.L
Source杭电ACM省赛集训队选拔赛之热身赛
RecommendEddy
查了再并...就是这么easy...
#include <set>#include <list>#include <cmath>#include <ctime>#include <deque>#include <queue>#include <stack>#include <cctype>#include <cstdio>#include <string>#include <vector>#include <cassert>#include <cstdlib>#include <cstring>#include <sstream>#include <iostream>#include <algorithm>using namespace std;int father[1001];int sum;int find(int x){if(x!=father[x])father[x] = find(father[x]);//回溯,所有节点都变为祖宗节点的子节点...即树的深度=2return father[x];}void Union(int a,int b){int x = find(a);int y = find(b);if(x==y)return;sum--;father[y]=x;}int main(){//freopen("in.txt","r",stdin);int x;cin>>x;while(x--){memset(father,0,sizeof(father));int n,m,i,j;cin>>n>>m;for(i=1;i<=1001;i++)father[i]=i;sum=n;while(m--){int a,b;cin>>a>>b;Union(a,b);}cout<<sum<<endl;} return 0;}