P3068 [USACO13JAN]派对邀请函Party Invitations

题目描述

Farmer John is throwing a party and wants to invite some of his cows to show them how much he cares about his herd. However, he also wants to invite the smallest possible number of cows, remembering all too well the disaster that resulted the last time he invited too many cows to a party.

Among FJ's cows, there are certain groups of friends that are hard to separate. For any such group (say, of size k), if FJ invites at least k-1 of the cows in the group to the party, then he must invite the final cow as well, thereby including the entire group. Groups can be of any size and may even overlap with each-other, although no two groups contain exactly the same set of members. The sum of all group sizes is at most 250,000.

Given the groups among FJ's cows, please determine the minimum number of cows FJ can invite to his party, if he decides that he must definitely start by inviting cow #1 (his cows are conveniently numbered 1..N, with N at most 1,000,000).

FJ正在举行派对,并想邀请他的一些奶牛参加以显示FJ多么关心他们,同时,他也希望邀请奶牛的数量最少,有了上一次派对的后果,他不要邀请过多的奶牛参加派对。

在FJ的奶牛中,有一些奶牛不能分开。 对于任何这样的奶牛群,(如果某个群的奶牛数量为k),当FJ邀请组中的至少k-1个奶牛参加派对时,他必须邀请最后的一头奶牛,从而包括整个组。 组可以是任何大小,并且甚至可以彼此重叠,但是没有两个组包含完全相同的奶牛。组的总数<=250,000。

考虑到FJ的牛群中的朋友关系群体的情况下,请你确定FJ可以邀请参加他的派对的奶牛的最小数量,开始前他必须先邀请编号为1的奶牛(为了方便,FJ的的奶牛方编号为1..N,N<=1,000,000)

输入输出格式

输入格式:

  • Line 1: Two space-separated integers: N (the number of cows), and G (the number of groups).

  • Lines 2..1+G: Each line describes a group of cows. It starts with an integer giving the size S of the group, followed by the S cows in the group (each an integer in the range 1..N).

第一行:两个空格分隔的整数:N和G。分别表示奶牛的数量和朋友组的的数量

接下来2到G+1行:每一行描述一群牛。它始于一个整数,表示这个组的大小,接下来每一个数表示一只奶牛的编号(编号在1到n之间,表示这只奶牛在这个朋友组里)

输出格式:

  • Line 1: The minimum number of cows FJ can invite to his party.

共一行,一个数:表示FJ最少可以邀请的牛的数量

输入输出样例

输入样例#1: 复制

10 4
2 1 3
2 3 4
6 1 2 3 4 6 7
4 4 3 2 1
输出样例#1: 复制

4

说明

There are 10 cows and 4 groups. The first group contains cows 1 and 3, and so on.

In addition to cow #1, FJ must invite cow #3 (due to the first group constraint), cow #4 (due to the second group constraint), and also cow #2 (due to the final group constraint).

样例一共10头牛和4组。 第一组包含奶牛1和3,接下来的组也类似

除了奶牛#1,FJ还必须邀请牛#3(由于第一组约束),牛#4(由于第二组约束)和牛#2(由于最终组约束)。

感谢mangoyang 提供翻译

#include<iostream>
#include<cstdio>
#include<set>
#include<queue>
#define maxn 1000001
using namespace std;
int num,head[maxn];
struct node{int to,pre;}e[maxn];
bool b[maxn];
queue<int>q;
set<int>s[maxn];
int n,m,x,y,ans;
void Insert(int from,int to){
e[++num].to=to;
e[num].pre=head[from];
head[from]=num;
}
int main(){
scanf("%d%d",&n,&m);
for(int i=;i<=m;i++){
scanf("%d",&x);
for(int j=;j<=x;j++){
scanf("%d",&y);
Insert(y,i);
s[i].insert(y);
}
}
q.push();
while(!q.empty()){
int now=q.front();q.pop();
for(int i=head[now];i;i=e[i].pre){
int to=e[i].to;
s[to].erase(now);
if(s[to].size()==){
set<int>::iterator it=s[to].begin();
if(!b[*it])b[*it]=,q.push(*it),ans++;
}
}
}
printf("%d",ans+);
}

洛谷P3068 [USACO13JAN]派对邀请函Party Invitations的更多相关文章

  1. 洛谷 P3068 [USACO13JAN]派对邀请函Party Invitations

    P3068 [USACO13JAN]派对邀请函Party Invitations 题目描述 Farmer John is throwing a party and wants to invite so ...

  2. [luoguP3068] [USACO13JAN]派对邀请函Party Invitations(stl大乱交)

    传送门 记录每一个编号在那些组中,可以用vector,这里选择链式前向星. 每一组用set 将被邀请的放到queue中 #include <set> #include <queue& ...

  3. [洛谷P3697]开心派对小火车

    题目:洛谷P3697 题目大意是有各站停列车(慢车,相邻2站时间A)和特急列车(相邻2站时间B),特急列车在特定站点停靠. 现在加一种快速列车(相邻2站时间C,A>C>B),停靠K站(包括 ...

  4. [洛谷201704R1]开心派对小火车

    OJ题号:洛谷P3697 思路: 贪心.首先从起点出发,开特急电车,对于每一个特急车站$s_{i}$,分别下一次车,计算从当前车站$s_{i}$出发坐各停电车在指定时限内$t$最远能够到达的车站$r_ ...

  5. 洛谷银牛派对SPFA

    题目描述 One cow from each of N farms (1 ≤ N ≤ 1000) conveniently numbered 1..N is going to attend the b ...

  6. 洛谷 P3071 [USACO13JAN]座位Seating-线段树区间合并(判断找,只需要最大前缀和最大后缀)+分治+贪心

    P3071 [USACO13JAN]座位Seating 题目描述 To earn some extra money, the cows have opened a restaurant in thei ...

  7. 洛谷P2202 [USACO13JAN]方块重叠Square Overlap

    P2202 [USACO13JAN]方块重叠Square Overlap 题目描述 Farmer John is planning to build N (2 <= N <= 50,000 ...

  8. 洛谷P3070 [USACO13JAN]岛游记Island Travels

    P3070 [USACO13JAN]岛游记Island Travels 题目描述 Farmer John has taken the cows to a vacation out on the oce ...

  9. 洛谷 P3071 [USACO13JAN]座位Seating(线段树)

    P3071 [USACO13JAN]座位Seating 题目链接 思路: 一开始把题给读错了浪费了好多时间呜呜呜. 因为第二个撤离操作是区间修改,所以我们可以想到用线段树来做.对于第一个操作,我们只需 ...

随机推荐

  1. 是否可以重定向到 WEB-INFO 下的页面?

    redirect的路径一定不能在WEB-INF路径下,因为redirect是相当于用户直接访问了路径,而用户不能访问WEB-INF目录下的文件,只有程序内部转发的时候才能转发到WEB-INF下的JSP ...

  2. Linux 基本命令总结

    0.平时自己用的命令 在跑一些程序,例如deep learning的时候,总是希望查看一下cpu.gpu.内存的使用率. (1)cpu.内存情况:使用top命令: (2)查看gpu:使用 nvidia ...

  3. [原创]Java开发在线编辑Word同时实现全文检索

    一.背景介绍 Word文档与日常办公密不可分,在实际应用中,当某一文档服务器中有很多Word文档,假如有成千上万个文档时,用户查找打开包含某些指定关键字的文档就变得很困难,一般情况下能想到的解决方案是 ...

  4. Git_错误_01_failed to push some refs to 'git@github.com

    在使用git 对源代码进行push到gitHub时可能会出错,信息如下 此时很多人会尝试下面的命令把当前分支代码上传到master分支上. $ git push -u origin master 但依 ...

  5. BEC listen and translation exercise 37

    You're supposed to do that before 10.30 in the morning, but obviously, if it's an emergency, you can ...

  6. FFMPEG内存操作(一) avio_reading.c 回调读取数据到内存解析

    相关博客列表 : FFMPEG内存操作(一) avio_reading.c 回调读取数据到内存解析 FFMPEG内存操作(二)从内存中读取数及数据格式的转换 FFmpeg内存操作(三)内存转码器 在F ...

  7. python密钥登录主机

    #!/usr/bin/python # -*- coding:utf-8 -*- ################################### # # 检查主机的损坏磁盘 # ####### ...

  8. BZOJ5362: [Lydsy1805月赛]quailty 算法

    BZOJ5362: [Lydsy1805月赛]quailty 算法 https://lydsy.com/JudgeOnline/problem.php?id=5362 分析: 题意即求一个最小基环树森 ...

  9. js检测对象属性

    In:(检测自身及原型属性) var o={x:1}; "x" in o; //true,自有属性存在 "y" in o; //false "toSt ...

  10. stm32之UCOS-III

    一.UCOS-III 学习UCOS-III,一般会学习以下内容: 任务创建.删除.挂起.恢复等: 临界区:独占CPU,尽量少用,否则会降低效率: 时间管理:时钟节拍(基于硬件定时器).软件定时器: 互 ...