哈理工oj hrbust 1725 Virus Outbreaking【并查集】

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Virus OutbreakingTime Limit: 1000 MSMemory Limit: 32768 KTotal Submit: 100(28 users)Total Accepted: 33(26 users)Rating: Special Judge: NoDescription

Village H is suffering a virus outbreaking.

There are n persons in the village, some of them is infected by at most one virus.

Input

There are multiple test cases. The first line is an integer T indicating the number of test cases.

The first line of each case is two positive integers n and q. n is the number of persons in the village and q is the times of event happened in village H.

The second line of each case is n numbers a1, a2, ..., an separated by space. ai stands for the kind of virus the ith person infected. 0 means not infected.

Then q lines following, each is an event happened in the village, consist of touching or query ordered by happening:

touch A B: a touched b and they infected each other. They will be together until the end.

query A: ask how many kind of virus person A infected. If A is not infected, output 0.

There will be a blank line after each case.

Note: 1 <= n <= 1000, 1 <= q <= 1000, 0 <= ai <= 32.

Output

For each query, output one line consist of all kinds of virus the person A infected, Output them by the increasing order, separated by space.

Output a blank line after each test case.

Sample Input

24 620 14 24 30query 1query 4query 4query 4touch 1 4query 24 67 4 28 20touch 4 1query 2query 1query 3query 3query 1

Sample Output

2030303014

47 2028287 20

Hint Source哈理工2013春季校赛 - 网络预选赛Author黄李龙@HRBUST

写完这最后一篇题解，就准备收拾收拾东西上火车了，寒假的最后一发、做的还算开心。

一道并查集的题目，英语好的，读懂题意马上就能AC；

题目大意：【自己看输出猜的】有n个人，m个提问，n个数代表n个人的权值，touch操作表示连接两个人，query表示查询这个人所在集合所有人的权值。【从小到大，并且不要重复】、

AC代码：

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;int a[1024];int f[1024];int n,m;int find(int x){    return f[x] == x ? x : (f[x] = find(f[x]));}void merge(int a,int b){    int A,B;    A=find(a);    B=find(b);    if(A!=B)    f[B]=A;}int findans(int b){    int ans[1024];    int cont=0;    for(int i=1;i<=n;i++)    {        if(find(i)==find(b))        {            int ok=1;            for(int j=0;j<cont;j++)            {                if(ans[j]==a[i])                {                    ok=0;                }            }            if(ok&&a[i])            {                ans[cont]=a[i];                cont++;            }        }    }    if(cont==0)    {        printf("0\n");    }    else    {        sort(ans,ans+cont);        for(int i=0;i<cont-1;i++)        {            printf("%d ",ans[i]);        }        printf("%d\n",ans[cont-1]);    }}int main(){    int t;    scanf("%d",&t);    while(t--)    {        scanf("%d%d",&n,&m);        for(int i=1;i<=n;i++)        {            scanf("%d",&a[i]);            f[i]=i;        }        while(m--)        {            char str[50];            scanf("%s",str);            if(str[0]=='t')            {                int x,y;                scanf("%d%d",&x,&y);                merge(x,y);            }            if(str[0]=='q')            {                int x;                scanf("%d",&x);                findans(x);            }        }        printf("\n");    }}

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