Cources

Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11298    Accepted Submission(s): 5299

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1083

Description:

Consider a group of N students and P courses. Each 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:

Sample Input:

2
3 3
3 1 2 3
2 1 2
1 1
3 3
2 1 3
2 1 3
1 1
Sample Output:
YES
NO
题意:
学生和课程一一匹配,问能不能做到最大匹配。
 
题解:
二分图匹配模板
代码如下:
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std; const int N = ;
int n,m,t,ans;
int check[N],match[N],link[N][N]; inline int dfs(int x){
for(int i=;i<=m;i++){
if(!check[i] && link[x][i]){
check[i]=;
if(match[i]==- || dfs(match[i])){
match[i]=x;
return ;
}
}
}
return ;
} int main(){
scanf("%d",&t);
while(t--){
scanf("%d%d",&n,&m);
ans = ;
memset(link,,sizeof(link));memset(match,-,sizeof(match));
for(int i=,x;i<=n;i++){
scanf("%d",&x);
for(int j=,k;j<=x;j++){
scanf("%d",&k);
link[i][k]=;
}
}
for(int i=;i<=n;i++){
memset(check,,sizeof(check));
if(dfs(i)) ans++;
}
if(ans==n) puts("YES");
else puts("NO");
}
return ;
}

HDU1083 :Courses(二分图匹配)的更多相关文章

  1. TZOJ 3030 Courses(二分图匹配)

    描述 Consider a group of N students and P courses. Each student visits zero, one or more than one cour ...

  2. HDU1083 Courses —— 二分图最大匹配

    题目链接:https://vjudge.net/problem/HDU-1083 Courses Time Limit: 20000/10000 MS (Java/Others)    Memory ...

  3. HDU-1083 Courses 二分图 最大匹配

    题目链接:https://cn.vjudge.net/problem/HDU-1083 题意 有一些学生,有一些课程 给出哪些学生可以学哪些课程,每个学生可以选多课,但只能做一个课程的代表 问所有课能 ...

  4. POJ1469 COURSES 二分图匹配 匈牙利算法

    原文链接http://www.cnblogs.com/zhouzhendong/p/8232649.html 题目传送门 - POJ1469 题意概括 在一个大矩阵中,有一些障碍点. 现在让你用1*2 ...

  5. POJ 1274 The Perfect Stall || POJ 1469 COURSES(zoj 1140)二分图匹配

    两题二分图匹配的题: 1.一个农民有n头牛和m个畜栏,对于每个畜栏,每头牛有不同喜好,有的想去,有的不想,对于给定的喜好表,你需要求出最大可以满足多少头牛的需求. 2.给你学生数和课程数,以及学生上的 ...

  6. Hdu1083 Courses

    Courses Problem Description Consider a group of N students and P courses. Each student visits zero, ...

  7. UVA 12549 - 二分图匹配

    题意:给定一个Y行X列的网格,网格种有重要位置和障碍物.要求用最少的机器人看守所有重要的位置,每个机器人放在一个格子里,面朝上下左右四个方向之一发出激光直到射到障碍物为止,沿途都是看守范围.机器人不会 ...

  8. POJ 1274 裸二分图匹配

    题意:每头奶牛都只愿意在她们喜欢的那些牛栏中产奶,告诉每头奶牛愿意产奶的牛棚编号,求出最多能分配到的牛栏的数量. 分析:直接二分图匹配: #include<stdio.h> #includ ...

  9. BZOJ1433 ZJOI2009 假期的宿舍 二分图匹配

    1433: [ZJOI2009]假期的宿舍 Time Limit: 10 Sec  Memory Limit: 162 MBSubmit: 2375  Solved: 1005[Submit][Sta ...

  10. HDU1281-棋盘游戏-二分图匹配

    先跑一个二分图匹配,然后一一删去匹配上的边,看能不能达到最大匹配数,不能这条边就是重要边 /*----------------------------------------------------- ...

随机推荐

  1. [SHELL]结构化命令之条件语句

    1.if-then语句  #!/bin/bash username="root" if grep $username /etc/passwd then echo "the ...

  2. springMVC怎么改变form的提交方式为put或者delete

    想着练习一下创建restful风格的网站呢,结果发现在jsp页面上并不能灵活使用put和delete提交方式.下面我的解决办法 一. form 只支持post和get两种提交方式,只支持get提交方式 ...

  3. 【转载】android 常用开源框架

    对于Android初学者以及对于我们菜鸟,这些大神们开发的轻量级框架非常有用(更别说开源的了). 下面转载这10个框架的介绍:(按顺序来吧没有什么排名). 一.  Afinal 官方介绍: Afina ...

  4. 创新手机游戏《3L》开发点滴(1)——道具、物品、装备表设计

    一.游戏物品/道具系统数据模型设计特点 为了让游戏更加的丰富,我们1201团队的新手机游戏设计了道具系统.于是丰富了游戏.取悦了玩家,哭了开发——道具/物品数据子系统是简单的.复杂的.不确定的: 简单 ...

  5. POJ 3415 Common Substrings(后缀数组)

    Description A substring of a string T is defined as: T(i, k)=TiTi+1...Ti+k-1, 1≤i≤i+k-1≤|T|. Given t ...

  6. UVALive 3668 A Funny Stone Game(博弈)

    Description The funny stone game is coming. There are n piles of stones, numbered with 0, 1, 2,...,  ...

  7. POJ 2449 Remmarguts' Date(第k短路のA*算法)

    Description "Good man never makes girls wait or breaks an appointment!" said the mandarin ...

  8. wpa_supplicant与kernel交互

    wpa_supplicant与kernel交互的操作,一般需要先明确驱动接口,以及用户态和kernel态的接口函数,以此来进行调用操作.这里分为4个步骤讨论. 1.首先需要明确指定的驱动接口.因为有较 ...

  9. 自测之Lesson16:并发通信

    知识点:三个多路并发模型(select .poll .epoll) 题目:以epoll模型,编写一个可供多个客户端访问的服务器程序. 实现代码: #include <netinet/in.h&g ...

  10. LintCode-112.删除排序链表中的重复元素

    删除排序链表中的重复元素 给定一个排序链表,删除所有重复的元素每个元素只留下一个. 样例 给出 1->1->2->null,返回 1->2->null 给出 1-> ...