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 \(m\) (\(1\le m\le n\le 5000\)), the number of students and the number of clubs in college.
The second line contains \(n\) integers \(p_1,p_2,…,p_n\) (\(0\le p_i<5000\)), where \(p_i\) is the potential of the \(i\)-th student.
The third line contains \(n\) integers \(c_1,c_2,…,c_n\) (\(1\le c_i\le m\)), which means that \(i\)-th student is initially a member of the club with index \(c_i\).
The fourth line contains an integer \(d\) (\(1\le d\le n\)), 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 \(k_i\) (\(1\le k_i\le n\)), 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\n",ans[i]);
return 0;
}
CF1139E Maximize Mex 题解【二分图】的更多相关文章
- CF1139E Maximize Mex(二分图匹配,匈牙利算法)
好题.不过之前做过的[SCOI2010]连续攻击游戏跟这题一个套路,我怎么没想到…… 题目链接:CF原网 洛谷 题目大意:在一个学校有 $n$ 个学生和 $m$ 个社团,每个学生有一个非负整数能力值 ...
- CF1139E Maximize Mex
题目地址:CF1139E Maximize Mex 这其实是一个二分图匹配匈牙利算法的巧妙运用 考虑倒序回答 则由原来的删除改为添加 把 potential 值作为左部,则一共有编号为 \(0~m\) ...
- Codeforces 1139E Maximize Mex 二分图匹配
Maximize Mex 离线之后把删数变成加数, 然后一边跑匈牙利一遍算答案. #include<bits/stdc++.h> #define LL long long #define ...
- [CF1139 E] Maximize Mex 解题报告 (二分图匹配)
interlinkage: https://codeforces.com/contest/1139/problem/E description: 有$n$个学生,$m$个社团,每个学生有一个能力值,属 ...
- codeforces1139E Maximize Mex 二分图匹配
题目传送门 题意:给出n个人,m个社团,每个人都有一个标号,一个能力值,并且属于一个社团,第i天的凌晨,第$k_i$个人会离开.每天每个社团最多派一个人出来参加活动.派出的人的能力值集合为S,求每天$ ...
- codeforces#1139E. Maximize Mex(逆处理,二分匹配)
题目链接: http://codeforces.com/contest/1139/problem/E 题意: 开始有$n$个同学和$m$,每个同学有一个天赋$p_{i}$和一个俱乐部$c_{i}$,然 ...
- BZOJ3339:Rmq Problem & BZOJ3585 & 洛谷4137:mex——题解
前者:https://www.lydsy.com/JudgeOnline/problem.php?id=3339 后者: https://www.lydsy.com/JudgeOnline/probl ...
- 洛谷4137 mex题解 主席树
题目链接 虽然可以用离线算法水过去,但如果强制在线不就gg了. 所以要用在线算法. 首先,所有大于n的数其实可以忽略,因为mex的值不可能大于n 我们来设想一下,假设已经求出了从0到n中所有数在原序列 ...
- 洛谷 P4137 Rmq Problem/mex 题解
题面 首先,由于本人太菜,不会莫队,所以先采用主席树的做法: 离散化是必须环节,否则动态开点线段数都救不了你: 我们对于每个元素i,插入到1~(i-1)的主席树中,第i颗线段树(权值线段树)对于一个区 ...
随机推荐
- tomcat启动了server.xml中没有配置的项目
在tomcat的conf目录下的server.xml文件中没有配置hczm_struts项目,但在eclipse启动tomcat调试时,一直启动hczm_struts项目. 经检查,发现conf\Ca ...
- css控制同一个页面的两个表格,一个显示有边框线,而另一个没边框线
不显示边框的:<table border="0" cellspacing="0" cellpadding="" ><tr ...
- Oracle学习笔记(八)
十一.子查询 1.子查询概述 学习子查询的原因 事例:查询工资比SCOTT高的员工信息 思路:1.scott的工资 select sal from emp where ename='SCOTT'; 2 ...
- swift 创建UICollectionView
// // CollectionViewController.swift // tab // // Created by su on 15/12/8. // Copyright © 2015年 ...
- 20169205实验四 Android程序设计
20169205实验四 Android程序设计 实验内容及步骤 (一)第一个android studio项目 android studio与eclipse的传统安卓开发有一些不同之处 android ...
- HDU1301&&POJ1251 Jungle Roads 2017-04-12 23:27 40人阅读 评论(0) 收藏
Jungle Roads Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 25993 Accepted: 12181 De ...
- B-spline Curves 学习之B样条曲线的系数计算与B样条曲线特例(6)
B-spline Curves: Computing the Coefficients 本博客转自前人的博客的翻译版本,前几章节是原来博主的翻译内容,但是后续章节博主不在提供翻译,后续章节我在完成相关 ...
- JSP和servlet之间的传值(总结的很全面)
转自:http://blog.csdn.net/ssy_shandong/article/details/9328985 1.从一个jsp页面跳转到另一个jsp页面时的参数传递 (1)使用re ...
- Objective-C 学习笔记(五) 指针
Objective-C 指针 每一个变量是一个内存位置和每一个存储单元都有其定义的地址,可以使用符号(&)的运算符,它表示内存中的地址访问. a. 我们定义一个指针变量 b. 分配一个指针变量 ...
- js练习计算器
js练习计算器,支持鼠标点击.键盘操作 <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" &quo ...