Mutual Training for Wannafly Union #6 E - Summer Trip（并查集）

最新推荐文章于 2022-12-09 19:15:36 发布

欧莎

最新推荐文章于 2022-12-09 19:15:36 发布

阅读量378

点赞数

CC 4.0 BY-SA版权

分类专栏： [kuangbin带你飞]专题五并查集

本文链接：https://blog.youkuaiyun.com/xiaosshhaa/article/details/60967071

[kuangbin带你飞]专题五并查集专栏收录该内容

8 篇文章

订阅专栏

本文介绍了一个基于并查集算法的学生旅行预算分配问题解决方案。一群学生计划前往海滩旅行，并需要合理分配预算。通过输入学生的数量、关系及各自的资金，算法能够找出不同群体并计算每个群体的总预算。

In this hot summer AIUB students decided to go on a trip to Cox’s Bazaar sea beach. Every student has some amount of money to spend in this trip. Each student belongs to some group. If student A knows student B, B knows C then A, B and C belong to the same group. In order to utilize the budget properly students hired a travel agency to get an optimal travel plan. They submitted the information about all the student’s money and relations among them. Now the agency wants to figure out the number of groups and the total budget of each group. Total budget of a group is the summation of all the student’s money in that group.

Input

The first line contains an integer T (1<=T<=50), denoting the number of test cases.

Each case starts with two integers N M. N (1<=N<=1000), denotes the number of students and M (0<=M<=(N*(N-1)/2)), denotes the number of relationships. In the next line N integers are given. The ith (1<=i<=N) integer a_i (1<=a_i<=100) denotes the money ith student has. Next M lines each contains two integers u v, (1<=u,v<=N and u != v) denoting that u knows v and vice versa.

Output

For each test case output “Case X: K” where X is the case number starting from 1 and K is the number of groups. In the next line output K integers, the total budget of all the groups in ascending order of budget.

Example

Input:
1
5 3
5 7 4 6 8
1 2
1 3
4 5

Output:
Case 1: 2

14 16

我比赛的时候用的dfs一直错，后来才知道为什么

#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>
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 pre[M];
int a[M];
int findset(int a)	//带路径压缩
{
    int r=a;
	while(pre[r] !=r)
	{
		r = pre[r];
	}
	int i=a,j;
    while(pre[i]!=r)
    {
        j=pre[i];
        pre[i]=r;
        i=j;
    }
    return r;
}

void union_nodes(int a, int b)     
{
	int a1 = findset(a);
	int b1 = findset(b);
	if(a1 != b1)		
	{
		pre[a1] = b1;		
	}
}
int main()
{
    int t,n,m,cnt=0;
    scanf("%d",&t);
    while(t--)
    {
        ++cnt;
        vector<int>ans,ss;
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            pre[i]=i;//初始化
        }
        for(int i=1;i<=m;i++)
        {
            int x,y;
            scanf("%d %d",&x,&y);
            union_nodes(x,y);
        }
        for(int i=1;i<=n;i++)
        {
            if(pre[i]==i) ss.push_back(i);
        }
        for(int i=0;i<ss.size();i++)
        {
            int sum=0;
            for(int j=1;j<=n;j++)
            {
                if(findset(j)==ss[i])
                    sum+=a[j];
            }
            ans.push_back(sum);
        }
        sort(ans.begin(),ans.end());
        int nn=ans.size();
        printf("Case %d: %d\n",cnt,nn);
        for(int i=0;i<nn;i++)
        printf("%d%c",ans[i],i==nn-1 ?'\n':' ');
    }
    return 0;
}