HUD——1083 Courses
HUD——1083 Courses
Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7699 Accepted Submission(s):
3781
student visits zero, one or more than one courses. Your task is to determine
whether it is possible to form a committee of exactly P students that satisfies
simultaneously the conditions:
. every student in the committee
represents a different course (a student can represent a course if he/she visits
that course)
. each course has a representative in the
committee
Your program should read sets of data from a text file. The
first line of the input file contains the number of the data sets. Each data set
is presented in the following format:
P N
Count1 Student1 1 Student1 2
... Student1 Count1
Count2 Student2 1 Student2 2 ... Student2
Count2
......
CountP StudentP 1 StudentP 2 ... StudentP CountP
The
first line in each data set contains two positive integers separated by one
blank: P (1 <= P <= 100) - the number of courses and N (1 <= N <=
300) - the number of students. The next P lines describe in sequence of the
courses . from course 1 to course P, each line describing a course. The
description of course i is a line that starts with an integer Count i (0 <=
Count i <= N) representing the number of students visiting course i. Next,
after a blank, you'll find the Count i students, visiting the course, each two
consecutive separated by one blank. Students are numbered with the positive
integers from 1 to N.
There are no blank lines between consecutive sets
of data. Input data are correct.
The result of the program is on the
standard output. For each input data set the program prints on a single line
"YES" if it is possible to form a committee and "NO" otherwise. There should not
be any leading blanks at the start of the line.
An example of program
input and output:
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#define N 501
using namespace std;
bool vis[N];
int n,m,x,y,T,t,ans,pre[N],map[N][N];
int read()
{
,f=;char ch=getchar();
; ch=getchar();}
+ch-'; ch=getchar();}
return x*f;
}
int find(int x)
{
;i<=n;i++)
{
if(!vis[i]&&map[x][i])
{
vis[i]=true;
||find(pre[i]))
{
pre[i]=x;
;
}
}
}
;
}
int main()
{
T=read();
while(T--)
{
ans=;
memset(map,,sizeof(map));
m=read(),n=read();
;i<=m;i++)
{
t=read();
while(t--)
{x=read();map[i][x]=;}
}
if(n<m) printf("NO\n");
else
{
memset(pre,-,sizeof(pre));
;i<=m;i++)
{
memset(vis,,sizeof(vis));
if(find(i)) ans++;
}
// printf("%d\n",ans);
if(ans==m) printf("YES\n");
else printf("NO\n");
}
}
;
}
HUD——1083 Courses的更多相关文章
- HDU 1083 Courses 【二分图完备匹配】
传送门:http://acm.hdu.edu.cn/showproblem.php?pid=1083 Courses Time Limit: 20000/10000 MS (Java/Others) ...
- HDU 1083 - Courses - [匈牙利算法模板题]
题目链接:http://acm.split.hdu.edu.cn/showproblem.php?pid=1083 Time Limit: 20000/10000 MS (Java/Others) M ...
- HDOJ 1083 Courses
Hopcroft-Karp算法模板 Courses Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 65536/32768 K (J ...
- hdoj 1083 Courses【匈牙利算法】
Courses Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total S ...
- hdu - 1083 - Courses
题意:有P门课程,N个学生,每门课程有一些学生选读,每个学生选读一些课程,问能否选出P个学生组成一个委员会,使得每个学生代言一门课程(他必需选读其代言的课程),每门课程都被一个学生代言(1 <= ...
- HDU - 1083 Courses /POJ - 1469
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1083 http://poj.org/problem?id=1469 题意:给你P个课程,并且给出每个课 ...
- HDU 1083 Courses(二分图匹配模板)
http://acm.hdu.edu.cn/showproblem.php?pid=1083 题意:有p门课和n个学生,每个学生都选了若干门课,每门课都要找一个同学来表演,且一个同学只能表演一门课,判 ...
- hdu 1083 Courses (最大匹配)
CoursesTime Limit: 20000/10000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Su ...
- HDU 1083 Courses(最大匹配模版题)
题目大意: 一共有N个学生跟P门课程,一个学生可以任意选一 门或多门课,问是否达成: 1.每个学生选的都是不同的课(即不能有两个学生选同一门课) 2.每门课都有一个代表(即P门课都被成功选过 ...
随机推荐
- C#调用dll(Java方法)
因为工作需求,要求用C#直接调用Java方法,下面呢是操作过程以及一些理解,如果有什么理解不对的,欢迎大家指出! 具体操作: 一.将Java写好的Demo以jar包形式导出 package demo; ...
- react基础语法(二)常用语法如:样式 ,自定义属性,常用表达式
<!DOCTYPE html> <html> <head> <meta charset="UTF-8"> <title> ...
- codevs 1422 河城荷取
时间限制: 1 s 空间限制: 128000 KB 题目等级 : 大师 Master 题目描述 Description 在幻想乡,河城荷取是擅长高科技工业的河童.荷取的得意之作除了光学迷彩外,还有 ...
- C++#pragma pack指令
微软官方文档说#pragma pack 指令的作用是为结构.联合和类成员指定 pack 对齐.的主要作用就是改变编译器的内存对齐方式,这个指令在网络报文的处理中有着重要的作用,#pragma pack ...
- pylint安装失败的解决方法
原文链接http://www.cnblogs.com/Loonger/p/7815335.html 使用命令pip3 install pylint安装pylint是出现错误.查了一圈也找不到答案.仔细 ...
- 目录下 shift 右键菜单 打开cmd 或者在 地址栏输入cmd 回车进入cmd
目录下 shift 右键菜单 打开cmd 或者在 地址栏输入cmd 回车进入cmd
- echart-柱状图
目前在改别人遗留的bug,需求: 宽度 自适应的情况下 展示不友好:宽度太大 上下不居中 需求 要 上下 无论是否 有内容 都要居中展示 以0刻度为标准 宽度 设置 series: [ { name: ...
- 搜索 || DFS || POJ 1321 棋盘问题
棋盘上#可以放,.不可以放,每行每列只能放一个 *解法:类似八皇后问题 dfs+回溯,考虑每一行和每一列 [[[[dfs的样子]]]]最前面写达到目标状态or不能走下去了 然后return #incl ...
- css内容补充之其它
1.overflow 当图片大小,超出div的大小时,可以指定overflow值为auto(带滚动条).hidden(隐藏,只显示一块): hover 当鼠标移动到当前标签上时,以下css属性才生效:
- 字符数组函数,连接strcat 复制函数strcpy 比较函数strcmp 长度函数 strlen
之前我们学习数据类型的时候,有一个类型 char ,这个类型允许我们在里边放一个字符 char variable1='o'; char variable2='k'; #include <iost ...