Hdu2860-Regroup(种类并查集)

Problem Description

When ALPC42 got to a panzer brigade, He was asked to build software to help them regroup the battalions or companies. As the tradition of army, soldiers are rated according his or her abilities, taking the rate as an integer. The fighting capacity of a company is determined by the soldier in this company whose rate is lowest. Now the recruits those rated are coming and join to their companies according to the order form HQ. With the coming of new recruits, a big regroup action reached, asking to merge some companies into one. The designation of a company, however, will not be canceled, but remain for memorialize what the company is done, means the designation of the company is still exist, but the company is gone, so it is unable to ask recruits to join this company, or merge the company into others. A strange thing is, the orders sometimes get wrong, send newbie to a company which is already merged into another, or mentioned some only-designation-existed companies. Such order could be rejected. The brigadier wants to know every change of each order, so the program should able to report the status of every order, telling whether it is accept, and can query the fighting capacity of specified company. (To simplify, companies are numbered from 0 to n-1

 

Input

There may be several test cases. For each case, the integers in first line, n, k, m, telling that there are n companies, k soldiers already, and m orders needs be executed. (1<=n ,k ,m<=100000). Then k lines with two integers R and C for each, telling a soldier with rate R is now in company C Then m lines followed, containing 3 kinds of orders, in upper case:   AP x y A recruit with ability rate x were asked to join company y. (0<=x<2^31, 0<=y<n)
  MG x y Company x and company y is merged. The new company is numbered as x. (0<=x, y<n)
  GT x Report the fighting capacity of company x. (0<=x<n)

 

Output

For each order there is exact one line to report the result. For AP and MG order, print “Accept” if it is able to be done, and execute it, or “Reject” if it is an illegal order. For GT order, if company x is still exist (not merged into others), print as “Lowest rate: y.” which y is the minimal rate of soldiers in this company. If there is no one in this company, tell "Company x is empty." If company x is already merged into others, print "Company x is a part of company z." z is the company where the company x is in. Print a blank line after each case

 

Sample Input

5 5 10

5 0

5 1

5 2

5 1

5 0

GT 0

GT 3

AP 3 3

GT 3

GT 4

MG 3 4

GT 4

MG 1 3

GT 4

GT 1

 

Sample Output

Lowest rate: 5.

Company 3 is empty.

Accept

Lowest rate: 3.

Company 4 is empty.

Accept

Company 4 is a part of company 3.

Accept

Company 4 is a part of company 1.

Lowest rate: 3.

 

题意：有n个公司，有k个人，会给出k个人的能力值和处于哪个公司。然后有m个操作，分别有3种不同的操作

AP x y 能力值为x的人加入到y公司中

MG x y x公司和y公司合并

GT x 查询x公司中能力值最小值

如果某个公司已经被别的公司合并了，则其他人不能加入此公司，此公司也不能再跟其他公司合并，对于AP和MG操作，如果可行则输出Accept，并执行此

操作，不能则输出Reject，忽略此操作。对于GT操作，如果此公司未被合并，则输出此公司中能力值最小值Lowest rate: 能力值. ，此公司没人则输出

Company 编号 is empty.否则输出Company 此公司编号 is a part of company 合并的公司编号.

 

解析：很明显的种类并查集，在合并的过程中更新一下能力值。这道题有一个坑点，如果x==y则直接输出Reject。。。。。。。。

 

#include<cstdio>
#include<cstring>
#include<string>
#include<iostream>
#include<sstream>
#include<algorithm>
#include<utility>
#include<vector>
#include<set>
#include<map>
#include<queue>
#include<cmath>
#include<iterator>
#include<stack>
using namespace std;
typedef __int64 LL;
const LL INF=1e12+7;
const int maxn=100005;
LL Min(LL a,LL b){ return a<b?a:b; }
int N,K,M;
LL val[maxn];
int d[maxn];
int root(int a)
{
    if(d[a]==a) return a;
    int t=d[a];
    d[a]=root(d[a]);
    //val[a]=Min(val[a],val[t]);
    return d[a];
}
void FunA()
{
    LL x;
    int y;
    scanf("%lld%d",&x,&y);
    int ra=root(y);
    if(ra!=y) printf("Reject
");
    else
    {
        printf("Accept
");
        val[ra]=Min(val[ra],x);
    }
}
void FunM()
{
    int x,y;
    scanf("%d%d",&x,&y);
    if(x==y){ printf("Reject
"); return; }
    int ra=root(x);
    int rb=root(y);
    if(ra!=x||rb!=y) printf("Reject
");
    else
    {
        d[rb]=ra;
        val[ra]=Min(val[ra],val[rb]);
        printf("Accept
");
    }
}
void FunG()
{
    int x;
    scanf("%d",&x);
    int ra=root(x);
    if(ra!=x) printf("Company %d is a part of company %d.
",x,ra);
    else if(val[ra]==INF) printf("Company %d is empty.
",ra);
    else printf("Lowest rate: %lld.
",val[ra]);
}
int main()
{
    while(scanf("%d%d%d",&N,&K,&M)!=EOF)
    {
        LL r;
        int c;
        for(int i=0;i<maxn;i++) val[i]=INF,d[i]=i;
        for(int i=1;i<=K;i++)
        {
            scanf("%lld%d",&r,&c);
            val[c]=Min(val[c],r);
        }
        char op[5];
        while(M--)
        {
            scanf("%s",op);
            if(op[0]=='A') FunA();
            else if(op[0]=='M') FunM();
            else FunG();
        }
        printf("
");
    }
    return 0;
}