Poj(1469),二分图最大匹配
题目链接:http://poj.org/problem?id=1469
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 21229 | Accepted: 8355 |
Description
- 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
Input
program should read sets of data from the std input. The first line of
the input 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抣l
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.
Output
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.
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
Source
给你p门课程和n个学生,一个学生可以选0门,1门,或者多门课程,现在要求一个由p个学生组成的集合,满足下列2个条件:
1.每个学生选择一个不同的课程
2.每个课程都有不同的代表
如果满足,就输出YES
分析:
做最大匹配,要是匹配数是P就是YES.
#include <stdio.h>
#include <string.h> int p,n;
bool maps[][];
bool use[];
int match[]; bool DFS(int u)
{
for(int i=; i<=n; i++)
{
if(!use[i]&&maps[u][i])
{
use[i] = true;
if(match[i]==-||DFS(match[i]))
{
match[i] = u;
return true;
}
}
}
return false;
} int main()
{
int t;
scanf("%d",&t);
while(t--)
{
memset(match,-,sizeof(match));
memset(maps,false,sizeof(maps));
scanf("%d%d",&p,&n);
int num = ;
for(int i=; i<=p; i++)
{
int stu;
scanf("%d",&stu);
for(int j=; j<=stu; j++)
{
int v;
scanf("%d",&v);
maps[i][v] = true;
}
} for(int i=; i<=p; i++)
{ memset(use,false,sizeof(use));
if(DFS(i))
num++;
}
if(num==p)
printf("YES\n");
else printf("NO\n"); } return ;
}
Poj(1469),二分图最大匹配的更多相关文章
- poj 1469 二分图最大匹配
就是最简单的最大匹配,没的说 #include<iostream> #include<cstdio> #include<cstring> #include<a ...
- poj 1469(二分图 最大匹配)
这道题让我认识到了c++cin,cout确实会使其超时,还是我用的c printf吧 #include<cstdio> #include<iostream> #include& ...
- poj 2239 二分图最大匹配,基础题
1.poj 2239 Selecting Courses 二分图最大匹配问题 2.总结:看到一个题解,直接用三维数组做的,很巧妙,很暴力.. 题意:N种课,给出时间,每种课在星期几的第几节课上 ...
- POJ 2226二分图最大匹配
匈牙利算法是由匈牙利数学家Edmonds于1965年提出,因而得名.匈牙利算法是基于Hall定理中充分性证明的思想,它是二部图匹配最常见的算法,该算法的核心就是寻找增广路径,它是一种用增广路径求二分图 ...
- POJ Evacuation /// 二分图最大匹配
题目大意: 在一个n*m的房间中 ‘X’为墙 ‘D’为门 ‘.’为人 门只存在与外围 人每秒钟只能向四连通区域走一步 门比较狭窄 每秒钟只能通过一个人 求所有人逃脱的最短时间 如果不可能则输出impo ...
- poj 2724 二分图最大匹配
题意: 会给出M个串,我们要做的就是将这M个串给清除了.对于任意两个串,若二进制形式只有一位不一样,那么这两个串可以在一次操作消除,否则每个操作只能消除一个串. 3 3 *01 100 011 可以代 ...
- Asteroids - poj 3041(二分图最大匹配问题)
Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 17258 Accepted: 9386 Description Be ...
- poj 2446 二分图最大匹配
思路:由(i+j)为偶数的点向(i+j)为奇数的点建边.求一次最大匹配,若正好为空格数(不包含洞)的一半,即输出YES. #include<iostream> #include<cs ...
- POJ 1719 二分图最大匹配(记录路径)
Shooting Contest Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 4097 Accepted: 1499 ...
- poj 3692 二分图最大匹配
思路: 如果我们将认识的建边,求最大独立集就是互相不认识的人数.那么我们反过来,将不认识的建图,求最大独立集就是互相认识的人数. #include<cstdio> #include< ...
随机推荐
- Leetcode: Insert Delete GetRandom O(1) - Duplicates allowed
Design a data structure that supports all following operations in average O(1) time. Note: Duplicate ...
- Lintcode: Majority Number II
Given an array of integers, the majority number is the number that occurs more than 1/3 of the size ...
- java-语句
JAVA语句 1.顺序语句(用:结束)(一个分号也是一个语句)(多条语句形成符合语句) 2.分支语句(又称条件语句) 1. if 语句 例: int a=10 if(a>0) {System ...
- Ubuntu 16.04 LTS Final Beta about JAVA
我们知道Ubuntu里可能会事先安装了openjdk.但是我习惯于Oracle jdk. ## < 卸载 openjdk > successfully: ### Terminal comm ...
- 事务——sql server中的事务应用举例
sql中事务只针对一个update,delete,insert语句,如果一段程序中有超过一个这样的语句,就需要每个都判断是否出错,否则就会出现若干我们不希望的情形出现,举例如下(表结构见最后): 1, ...
- Thread create 创建进程
#include "windows.h" #include "iostream" #include "stdio.h" void Start ...
- C 排序法
1.冒泡法,相邻的两个数值,进行比较,满足条件的进行互换 #include <stdio.h> int main() { int index, j, tmp; , , ,}; ; inde ...
- -05 08:57 ARCGIS地统计学计算文件后缀名为.shp文件制作
2011-07-05 08:57 ARCGIS地统计学计算文件后缀名为.shp文件制作 ARCAMP软件要进行地统计计算的文件后格式后缀名必须为.shp的文件,网上介绍的方法复杂难懂,那么制作.shp ...
- DataTable 筛选数据
//使用聚合函数 max ,sum ,count .... private void ComputeBySalesSalesID(DataSet dataSet) { // Presumes ...
- oracle的会话(session)
会话(session)是oracle服务器对数据库连接用户记录的一种手段. oracle提供了v_$session的视图存储当前数据库的会话,查询时用v_$session 或v$session sql ...