CF1139E Maximize Mex 题解【二分图】

我发现我有道叫[SCOI2010]连续攻击游戏的题白写了..

Description

There are (n) students and (m) clubs in a college. The clubs are numbered from (1) to (m). Each student has a potential (p_i) and is a member of the club with index (c_i). Initially, each student is a member of exactly one club. A technical fest starts in the college, and it will run for the next (d) days. There is a coding competition every day in the technical fest.

Every day, in the morning, exactly one student of the college leaves their club. Once a student leaves their club, they will never join any club again. Every day, in the afternoon, the director of the college will select one student from each club (in case some club has no members, nobody is selected from that club) to form a team for this day's coding competition. The strength of a team is the mex of potentials of the students in the team. The director wants to know the maximum possible strength of the team for each of the coming dddays. Thus, every day the director chooses such team, that the team strength is maximized.

The mex of the multiset (S) is the smallest non-negative integer that is not present in (S). For example, the mex of the ({0,1,1,2,4,5,9}) is (3), the mex of ({1,2,3}) is (0) and the mex of (varnothing) (empty set) is (0).

Input

The first line contains two integers (n) and (m) ((1le mle nle 5000)), the number of students and the number of clubs in college.

The second line contains (n) integers (p_1,p_2,…,p_n) ((0le p_i<5000)), where (p_i) is the potential of the (i)-th student.

The third line contains (n) integers (c_1,c_2,…,c_n) ((1le c_ile m)), which means that (i)-th student is initially a member of the club with index (c_i).

The fourth line contains an integer (d) ((1le dle n)), number of days for which the director wants to know the maximum possible strength of the team.

Each of the next (d) lines contains an integer (k_i) ((1le k_ile n)), which means that (k_i)-th student lefts their club on the (i)-th day. It is guaranteed, that the (k_i)-th student has not left their club earlier.

Output

For each of the (d) days, print the maximum possible strength of the team on that day.

Examples

input

5 3
0 1 2 2 0
1 2 2 3 2
5
3
2
4
5
1

output

3
1
1
1
0

input

5 3
0 1 2 2 1
1 3 2 3 2
5
4
2
3
5
1

output

3
2
2
1
0

input

5 5
0 1 2 4 5
1 2 3 4 5
4
2
3
5
4

output

1
1
1
1

Note

Consider the first example:

On the first day, student (3) leaves their club. Now, the remaining students are (1), (2), (4) and (5). We can select students (1), (2) and (4) to get maximum possible strength, which is (3). Note, that we can't select students (1), (2) and (5), as students (2) and (5) belong to the same club. Also, we can't select students (1), (3) and (4), since student (3) has left their club.

On the second day, student (2) leaves their club. Now, the remaining students are (1), (4) and (5). We can select students (1), (4) and (5) to get maximum possible strength, which is (1).

On the third day, the remaining students are (1) and (5). We can select students (1) and (5) to get maximum possible strength, which is (1).

On the fourth day, the remaining student is (1). We can select student (1) to get maximum possible strength, which is (1).

On the fifth day, no club has students and so the maximum possible strength is (0).

题意

有 (n) 个学生，(m) 个社团。每个学生属于一个社团。在接下来的 (d) 天里，每天会有一个人退团。每天从每个社团中最多选出一个人，使得选出的人的能力值集合 ({p_i}) 的 (mathrm{mex}) 值最大。求出每天的最大 (mathrm{mex}) 值。

题解

可能建立二分图模型是求 (mathrm{mex}) 的一个套路/技巧吧。反正是忘了…但是做题的时候也不知道哪些是经典，干脆以后全部好好落实。

因为每个能力值至少被一个社团提供，而一个社团最多提供一个能力值，由此构造二分图。左侧是能力值，右侧是社团。当存在一个学生能力值为 (p_i)，社团为 (c_i) 时，从 (p_i) 向 (c_i) 连边。

但是学生是在动态变化的，而且只会减少人，由此每次询问的答案一定非严格递减。

但是二分图最大匹配的匈牙利算法是不支持删边的，而一次匹配（指左边的一个点）的复杂度就是 (O(m))，每次进行复杂度高达 (O(nmd)​)，显然不是我们想要的。

考虑反过来，从最终状态加学生，每次的答案一定不会减少。那么我们就假设加完前 (i) 个学生后，此时的 (mathrm{mex}) 是 (t)。在二分图左侧 Find(t)，此时如果替换，只会替换比 (t) 小的点，不会造成更劣答案；最终会找到一个没有匹配过的社团，令答案 (+1)。

重新加边不会影响原来算好的答案。最终反序输出即可。

时间复杂度 (O(nm+qm))。

Code：

写匈牙利一定要在每次 Find() 函数外调用 Find() 时清空 used[] 啊啊啊啊啊啊否则会 FST

#include<cstdio>
#include<cstring>
struct edge
{
    int n,nxt;
    edge(int n,int nxt)
    {
        this->n=n;
        this->nxt=nxt;
    }
    edge(){}
}e[5050];
int head[5050],ecnt=-1;
void add(int from,int to)
{
    e[++ecnt]=edge(to,head[from]);
    head[from]=ecnt;
}
int s[5050];
bool used[5050];
bool Find(int x)
{
    if(used[x])
        return false;
    used[x]=1;
    for(int i=head[x];~i;i=e[i].nxt)
        if(s[e[i].n]==-1||Find(s[e[i].n]))
        {
            s[e[i].n]=x;
            return true;
        }
    return false;
}
int a[5050],b[5050],c[5050];
int ans[5050];
int main()
{
    memset(head,-1,sizeof(head));
    memset(s,-1,sizeof(s));
    int n,m,q;
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;++i)
        scanf("%d",&a[i]);
    for(int i=1;i<=n;++i)
        scanf("%d",&b[i]);
    scanf("%d",&q);
    for(int i=1;i<=q;++i)
    {
        scanf("%d",&c[i]);
        used[c[i]]=1;
    }
    for(int i=1;i<=n;++i)
        if(!used[i])
            add(a[i],b[i]);
    int t=0;
    for(int i=q;i>=1;--i)
    {
        memset(used,0,sizeof(used));
        while(Find(t))
        {
            ++t;
            memset(used,0,sizeof(used));
        }
        ans[i]=t;
        add(a[c[i]],b[c[i]]);
    }
    for(int i=1;i<=q;++i)
        printf("%d
",ans[i]);
    return 0;
}