Summer Trip

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

题意：给你n个人编号从1~n，每个人有一定的价钱，然后有m个关系，有关系的人可以分到一组，问你能分几组，并且把每组的价钱输出来

思路：并查集，只要每个人的父节点一样就可以分到一组

#include<stdio.h>#include<iostream>#include<string.h>#include<algorithm>using namespace std;#define N 10010int father[N];int vis[10100];int a[100010];int s[10010];int we[10110],ni[10100];int ha[100010];int getfather(int x)//找父节点{    while(x!=father[x])    {        x=father[x];    }    return x;}void un(int x,int y)//更新父节点{    int x1=getfather(x);    int y1=getfather(y);    if(x1!=y1)        father[x1]=y1;}bool same(int x,int y)//判断父节点是否相同{    return getfather(x)==getfather(y);}int main(){    int g=0;    int t;    scanf("%d",&t);    int n,m;    while(t--)    {        memset(ha,0,sizeof(ha));        memset(s,0,sizeof(s));        memset(vis,0,sizeof(vis));        scanf("%d%d",&n,&m);        ++g;        memset(father,0,sizeof(father));        for(int i=1; i<=n; i++)        {            father[i]=i;        }        for(int i=1; i<=n; i++)        {            scanf("%d",&a[i]);        }        for(int i=1; i<=m; i++)        {            scanf("%d%d",&we[i],&ni[i]);            un(we[i],ni[i]);        }        for(int i=1;i<=n;i++)        {            ha[getfather(i)]+=a[i];        }        int p=0;        for(int i=1;i<=n;i++)        {            if(ha[i]!=0)            {                s[p++]=ha[i];            }        }        printf("Case %d: %d\n",g,p);        sort(s,s+p);        for(int i=0; i<p; i++)        {            if(s[i]&&i<n)                printf("%d ",s[i]);                if(s[i]&&i==n)                    printf("%d\n",s[i]);        }    }}