HDU1213 并查集

How Many Tables

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 20906    Accepted Submission(s): 10350

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N个男孩在一个房间里面，这个叫XX的人指定其中几个人为好朋友，好朋友能相互传递，比如说A和B是好朋友,AC也是好朋友，那么BC就是好朋友。求，最长的那一堆好朋友有几个人

遇到的问题和解决思路：

首先就是并查集问题。这里可以运用树根是多少，把所有的人都排到一个人上面去就可以了。当然rank不能设置成0，如果设置能0，那么不管怎么加都是0。然后设置成1的原理就是，因为每个人刚开始就是一个单位，所以刚开始的根就是1.

还有就是n可能是0，也就是没有一个人拥有朋友。

给出代码：

这里的原则是，把所有的根上的值都加到顶端上去。单纯的利用par[x] = y; rank[y] += rank[x]

#include<cstdio>
#include<algorithm>
#include<cstring>

using namespace std;


const int MAX_N = 100000 + 20;
const int MAX_R = 10000000 + 50;
int par[MAX_R], rank1[MAX_R];
int n, high, a, b;

void init(){
	for(int i = 1; i <= MAX_R - 30; i++){
		par[i] = i;
		rank1[i] = 1;//rank一定要是1
	}
}

int find(int x){
	if(par[x] == x)return x;
	return par[x] = find(par[x]);
}

bool same(int x, int y){
	return find(x) == find(y);
}

void unite(int x, int y){
	x = find(x);
	y = find(y);
	if(x == y) return ;
	par[x] = y;
	rank1[y] += rank1[x];
}


int main(){
	while(scanf("%d", &n) != EOF){
		if(n == 0){
			printf("%d\n", 1);
			continue;
		}
		init();
		high = 0;
		for(int i = 1; i <= n; i++){
			scanf("%d%d", &a, &b);
			if(!same(a, b)){
				unite(a, b);
				high = max(high, max(rank1[find(a)], rank1[find(b)]));//寻找最大的
			}
		}
		printf("%d\n", high);
	}
	return 0;
}

后来我又思考了一下，把小的当成根，把大的当做末尾，然后每一次就是+1，但是因为刚开始少加了1，所以最后输出high+1.

给出代码（就只有unite和printf这里有不同）：

这里的是，把所有的值放到末尾上去。

#include<cstdio>
#include<algorithm>
#include<cstring>
 using namespace std; const int MAX_N = 100000 + 20; const int MAX_R = 10000000 + 50; int par[MAX_R], rank1[MAX_R]; int n, high, a, b; void init(){ for(int i = 1; i <= MAX_R - 30; i++){
        par[i] = i;
        rank1[i] = 0; } } int find(int x){ if(par[x] == x)return x; return par[x] = find(par[x]); } bool same(int x, int y){ return find(x) == find(y); } void unite(int x, int y){
    x = find(x);
    y = find(y); if(x == y) return ; if(x > y){
        par[y] = x;
        rank1[x] = rank1[x] + rank1[y] + 1; } else {
        par[x] = y;
        rank1[y] = rank1[x] + rank1[y] + 1; } } int main(){ while(scanf("%d", &n) != EOF){ if(n == 0){
            printf("%d\n", 1); continue; }
        init();
        high = 0; for(int i = 1; i <= n; i++){
            scanf("%d%d", &a, &b); if(!same(a, b)){
                unite(a, b);
                high = max(high, max(rank1[find(a)], rank1[find(b)])); } }
        printf("%d\n", high + 1); } return 0; }