SPOJ：Lost and survived（multiset+并查集）

Posted 2020-11-02 ---学习ing---

On September 22, 2004, Oceanic Flight 815 crashed on a mysterious island somewhere in the pacific.

There actually were survivors in the crash , N survivors . The mysterious island kept on moving in space - time, so no rescue reached them.

Initially every survivor was on his own .But soon they realized there are these “The Others” (Scientists of damn Dharma Initiative) on this Island too.

So to protect themselves from mad Scientists they started uniting into groups after Dr. Shephard  said  “ Live together Die alone ”.

You have to handle Q queries; which consist of two survivors becoming friends and thereby uniting there respective groups into a  larger group.

After each query, output the difference between the group of largest size and group of smallest size at that time.

If there is only one group, output 0. At first, everyone is in their own group.

Also note, if the two survivors in the query are already in the  same group, print the current answer, and skip merging groups.

Also do comment if you have watched Lost :-p

Input

 

The first line consists of two space separated integers, N and Q

The next Q line consists of two integers, A and B, meaning that 

the groups involving survivor A and survivor B unite into a larger group.

The first line consists of two space separated integers, N and Q

 

The next Q line consists of two integers, A and B, meaning that 

survivor A and survivor B became friends uniting there groups.

 

Output

Output Q lines, the answer after each query.

 

1<=N<=100000

1<=Q<=100000

Example

Input:
5 3
1 2
2 3
5 4

Output:
1
2
1

 Submit solution!

 题意：有N个人，一开始都自己一个集合。有Q次操作，每次给定u，v，要求合并u和v到一个集合（已经在就忽略），然后输出目前的最大集合元素个数减去最小集合元素个数。

思路：集合用并查集表示，元素个数用multiset保存。

注意multiset的删除：s.erase(find(x))，只删除一个；而 s.erase(x)， 会删除所有值为x的元素    

#include<bits/stdc++.h>
using namespace std;
const int maxn=100010;
multiset<int>s;
int fa[maxn],num[maxn],Max=1,Min=1;
int find(int u){
    if(u==fa[u]) return u;
    fa[u]=find(fa[u]); return fa[u];
}
void merge(int u,int v){
    int fau=find(u),fav=find(v);
    if(num[fau]>num[fav]) swap(fau,fav);
    fa[fau]=fav;
    s.erase(s.find(num[fau])); s.erase(s.find(num[fav]));
    num[fav]+=num[fau];    s.insert(num[fav]);  
    if(num[fav]>Max) Max=num[fav];
    Min=*s.begin();
}
int main()
{
    int N,Q,u,v,i,j,ans;
    scanf("%d%d",&N,&Q);
    for(i=1;i<=N;i++){
        fa[i]=i; num[i]=1;
        s.insert(1);
    }
    while(Q--){
        scanf("%d%d",&u,&v);
        if(find(u)!=find(v))  merge(u,v);
        printf("%d\n",Max-Min);
    }
    return 0;
}