poj——1469 COURSES
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 24192 | Accepted: 9426 |
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
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
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
题目大意:一些课一些人,组成一个集体,集体中每人代表每门不同的课,每门课在集体中有一名成员,问是否能组成这样的集体
思路:
这个题我们只需要求出二分图的最大匹配数,这样我们求出他们选的课程数,然后判断最大匹配数是否等于课程总数。
代码:
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#define N 501
using namespace std;
bool vis[N];
int t,w,n,m,x,ans,girl[N],map[N][N];
int read()
{
,f=; char ch=getchar();
; ch=getchar();}
+ch-';ch=getchar();}
return x*f;
}
int find(int x)
{
;i<=m;i++)
{
if(!vis[i]&&map[x][i])
{
vis[i]=true;
||find(girl[i])) {girl[i]=x;;}
}
}
;
}
int main()
{
t=read();
while(t--)
{
ans=;
memset(map,,sizeof(map));
n=read(),m=read();
;i<=n;i++)
{
w=read();
;}
}
if(m<n){printf("NO\n"); continue;}
memset(girl,-,sizeof(girl));
;i<=n;i++)
{
memset(vis,,sizeof(vis));
if(find(i)) ans++;
}
if(ans==n) printf("YES\n");
else printf("NO\n");
}
;
}
poj——1469 COURSES的更多相关文章
- 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 ...
- poj 1469 COURSES(匈牙利算法模板)
http://poj.org/problem?id=1469 COURSES Time Limit: 1000MS Memory Limit: 10000K Total Submissions: ...
- POJ 1469 COURSES 二分图最大匹配 二分图
http://poj.org/problem?id=1469 这道题我绝壁写过但是以前没有mark过二分图最大匹配的代码mark一下. 匈牙利 O(mn) #include<cstdio> ...
- poj 1469 COURSES 解题报告
题目链接:http://poj.org/problem?id=1469 题目意思:有 N 个人,P个课程,每一个课程有一些学生参加(0个.1个或多个参加).问 能否使得 P 个课程 恰好与 P 个学生 ...
- POJ 1469 COURSES
COURSES Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 20478 Accepted: 8056 Descript ...
- POJ 1469 COURSES(二部图匹配)
COURSES Time Limit: 1000MS Memory ...
- poj 1469 COURSES 题解
COURSES Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 21515 Accepted: 8455 Descript ...
- poj 1469 COURSES (二分匹配)
COURSES Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 16877 Accepted: 6627 Descript ...
- poj 1469 COURSES (二分图模板应用 【*模板】 )
COURSES Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 18454 Accepted: 7275 Descript ...
随机推荐
- 【JavaScript】随机生成10个0~100的数字
随机生成10个0~100不重复的数字(包含0和100): 需要用到的知识点:随机数 去重 下面放代码 <!DOCTYPE html> <html> <head> & ...
- HttpWebRequest 以及WebRequest的使用
1.WebRequest的发送数据以及接收数据 class Program { static void Main(string[] args) { //创建一个实例并发送请求 HttpWebReque ...
- DELL笔记本安装Ubuntu 14.04
1. 将制作好的USB启动盘插入电脑,按f2进入启动选择选项,选择U盘启动: 进入选择界面后讲光标移动到"install Ubuntu"选项,按'e'进入grub界面,将倒数第二行 ...
- mybatis-paginator对SqlServer分页实现
package com.github.miemiedev.mybatis.paginator.dialect; import com.github.miemiedev.mybatis.paginato ...
- 翻译:高级t - sql第1级的阶梯:使用交叉连接来引入高级t - sql
高级t - sql第1级的阶梯:使用交叉连接来引入高级t - sql 源于:格雷戈里·拉森,2016/02/19(首次出版:2014/12/17 翻译:刘琼滨 谢雪妮 徐雅莉 赖慧芳 链接:http: ...
- Node.js——重定向
- android 设置跳转
android.provider.Settings. 1. ACTION_ACCESSIBILITY_SETTINGS : // 跳转系统的辅助功能界面 Intent ...
- WNDCLASS和WNDCLASSEX
typedef struct { UINT cbSize; UINT style; WNDPROC lpfnWndProc; int cbClsExtra; int cbWndExtra; HINST ...
- (独孤九剑)--PHP简介与现况
(1)为什么学习PHP? 1.好就业: 2.入门简单,学习周期短,两个月即可: 3.学习编程思路,使编程习惯更加规范: 4.大公司直招: 5.处理大并发数据: 6.开源,所以更加安全 (2)PHP是什 ...
- 01XML文档结构
文档结构 2.1文档结构 2.1.1文档声明及字符编码 <?xml version=“1.0” encoding=“”gb2312 standalone=“yes”?> <? 告诉 ...