【C - How Many Tables】

本文深入探讨了并查集算法的实现与应用,通过求解联通块数量的问题,介绍了并查集的基本操作,如初始化、查找和合并。并提供了一个0ms运行的C++代码示例,展示了算法的高效性和简洁性。

思路:

  • 非常质朴的并查集啦,求有几个联通块,最后头为 -1 的元素个数就是呗。

代码:

  • 0ms 1416kB
//0ms		1416kB


#include <iostream>
#include <cstring>
#include <cstdio>

using namespace std;

const int maxn = 1005;

int N,M;
int ans;
int par[maxn];

void INIT(){
	memset(par , -1 , sizeof(par));
	ans = 0;
	return ;
}

int FIND(int i){
	return par[i] == -1 ? i : par[i] = FIND(par[i]);
}

void UNION(int l,int r){
	int parl = FIND(l);
	int parr = FIND(r);
	if(parr != parl)
		par[parr] = parl;
	return ;
}

int main(){
	int T;cin>>T;
	while(T--){
		INIT();
		scanf("%d%d" , &N , &M);
		while(M--){
			int l,r;
			scanf("%d%d" , &l , &r);
			UNION(l , r);
		}
		for(int i=1;i<=N;i++)
			if(par[i] == -1)
				ans++;
		printf("%d\n" , ans);
	}
	return 0;
}
6-4 朋友聚会 分数 10 作者 杜祥军 单位 青岛大学 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. 函数接口定义: int find(int x); 其中 N 和 D 都是用户传入的参数。 N 的值不超过int的范围; D 是[0, 9]区间内的个位数。函数须返回 N 中 D 出现的次数。 裁判测试程序样例: #include <stdio.h> int pre[1010]; int find(int x); int main() { int t; scanf("%d",&t); while(t--){ int n ,m; scanf("%d%d",&n,&m); for(int i=1;i<=n;i++) pre[i]=i; for(int i=0;i<m;i++){ int x,y; scanf("%d%d",&x,&y); int fx=find(x); int fy=find(y); if(fx!=fy) pre[fx]=fy; } int cnt=0; for(int i=1;i<=n;i++) if(pre[i]==i) cnt++; printf("%d\n",cnt); } return 0; } /* 请在这里填写答案 */ 输入样例: 2 5 3 1 2 2 3 4 5 5 1 2 5 输出样例: 2 4 代码长度限制 16 KB 时间限制 1000 ms 内存限制 32 MB C++ (g++) 1 给出参考代码
最新发布
10-28
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