【CF1037E】Trips

题目描述

There are $n$ persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.

We want to plan a trip for every evening of $m$ days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:

Either this person does not go on the trip,
Or at least $k$ of his friends also go on the trip.

Note that the friendship is not transitive. That is, if $a$ and $b$ are friends and $b$ and $c$ are friends, it does not necessarily imply that $a$ and $c$ are friends.

For each day, find the maximum number of people that can go on the trip on that day.

输入输出格式

输入格式：

The first line contains three integers $n$ , $m$ , and $k$ ($ 2≤n≤210^5,1≤m≤210^5 ,1≤k<n $) — the number of people, the number of days and the number of friends each person on the trip should have in the group.

The $i -th$ ( $1≤i≤m$ ) of the next $m$ lines contains two integers $x$ and $y$ ( $1≤x,y≤n , x≠y $ ), meaning that persons $x$and $y$ become friends on the morning of day $i$ . It is guaranteed that $x$ and $y$ were not friends before.

输出格式：

Print exactly $m$ lines, where the $i -th$ of them ( $1≤i≤m$ )contains the maximum number of people that can go on the trip on the evening of the day $i$ .

输入输出样例

输入样例#1：

输出样例#1：

输入样例#2：

输出样例#2：

输入样例#3：

输出样例#3：

分析

考虑对朋友之间建边，这样便将问题转化为动态加边了。然后每天选出一些点，使这些点与其他被选的点有至少 $k$ 条连边。但正着做不好做，于是考虑反着做。

我们先将所有的边都连上，如果这时点的度数小于 $k$ ，那么这个点无论如何也不可能被选，所以我们直接将其删除，同时更新与其相邻的点。此时剩余的点即第 $m$ 天的答案。

最后，倒着删边，统计答案即可。

代码

#include<set>
#include<queue>
#include<cstdio>
#include<cstring>
#include<iostream>
#define il inline
#define re register
#define maxn 200005
#define tie0 cin.tie(0),cout.tie(0)
#define fastio ios::sync_with_stdio(false)
using namespace std;
typedef long long ll;

template<typename T>inline void read(T &x){
    T f=1;x=0;char c;
    for(c=getchar();!isdigit(c);c=getchar())if(c=='-')f=-1;
    for(;isdigit(c);c=getchar())x=x*10+(c^48);
    x*=f;
}

struct edge{
    int u,v;
}e[maxn];

set<int>st[maxn];

int n,m,k,cnt;
int ans[maxn],du[maxn];
bool del[maxn];

void Delete(int x){
    if(du[x]>=k||del[x]) return;
    del[x]=1;cnt--;
    queue<int>q;
    q.push(x);
    while(!q.empty()){
        int now=q.front();
        q.pop();
        for(auto y : st[now]){
            du[y]--;
            if(du[y]<k&&!del[y]){
                q.push(y);
                del[y]=1;cnt--;
            }
        }
    }
}

int main(){
    int x,y;
    read(n),read(m),read(k);
    cnt=n;
    for(int i=1;i<=m;++i){
        read(x),read(y);
        e[i].u=x,e[i].v=y;
        du[x]++,du[y]++;
        st[x].insert(y),st[y].insert(x);
    }
    for(int i=1;i<=n;++i)
        Delete(i);
    ans[m]=cnt;
    for(int i=m;i;--i){
        x=e[i].u,y=e[i].v;
        if(!del[x]) du[y]--;
        if(!del[y]) du[x]--;
        st[x].erase(y),st[y].erase(x);
        Delete(x),Delete(y);
        ans[i-1]=cnt;
    }
    for(int i=1;i<=m;++i) printf("%d\n",ans[i]);
    return 0;