POJ2239_Selecting Courses(二分图最大匹配)

解题报告

http://blog.csdn.net/juncoder/article/details/38154699

题目传送门

题意:

每天有12节课。一周上7天,一门课在一周有多天上课。

求一周最多上几节课。

思路:

把课程看成一个集合,上课的时间看成一个集合,二分图就出来了。

#include <cstdio>
#include <cstring>
#include <iostream>

using namespace std;
int n,day[10][15],mmap[500][500],vis[500],cnt,pre[500];
int dfs(int x)
{
    for(int i=n+1;i<cnt;i++){
        if(!vis[i]&&mmap[x][i]){
            vis[i]=1;
            if(pre[i]==-1||dfs(pre[i])){
                pre[i]=x;
                return 1;
            }
        }
    }
    return 0;
}
int main()
{
    int i,j,a,b,t;
    while(cin>>n){
        cnt=n+1;
        memset(mmap,0,sizeof(mmap));
        memset(day,0,sizeof(day));
        memset(pre,-1,sizeof(pre));
        for(i=1;i<=n;i++){
            cin>>t;
            for(j=1;j<=t;j++){
                cin>>a>>b;
                if(!day[a][b])
                {
                    day[a][b]=cnt++;
                }
                mmap[i][day[a][b]]=1;
            }
        }
        int ans=0;
        for(i=1;i<=n;i++){
            memset(vis,0,sizeof(vis));
            ans+=dfs(i);
        }
        cout<<ans<<endl;
    }
    return 0;
}

Selecting Courses

Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 8466   Accepted: 3769

Description

It is well known that it is not easy to select courses in the college, for there is usually conflict among the time of the courses. Li Ming is a student who loves study every much, and at the beginning of each term, he always wants to select courses as more
as possible. Of course there should be no conflict among the courses he selects.

There are 12 classes every day, and 7 days every week. There are hundreds of courses in the college, and teaching a course needs one class each week. To give students more convenience, though teaching a course needs only one class, a course will be taught several
times in a week. For example, a course may be taught both at the 7-th class on Tuesday and 12-th class on Wednesday, you should assume that there is no difference between the two classes, and that students can select any class to go. At the different weeks,
a student can even go to different class as his wish. Because there are so many courses in the college, selecting courses is not an easy job for Li Ming. As his good friends, can you help him?

Input

The input contains several cases. For each case, the first line contains an integer n (1 <= n <= 300), the number of courses in Li Ming‘s college. The following n lines represent n different courses. In each line, the first number is an integer t (1 <= t <=
7*12), the different time when students can go to study the course. Then come t pairs of integers p (1 <= p <= 7) and q (1 <= q <= 12), which mean that the course will be taught at the q-th class on the p-th day of a week.

Output

For each test case, output one integer, which is the maximum number of courses Li Ming can select.

Sample Input

5
1 1 1
2 1 1 2 2
1 2 2
2 3 2 3 3
1 3 3

Sample Output

4
时间: 2024-10-13 16:04:44

POJ2239_Selecting Courses(二分图最大匹配)的相关文章

poj2239 Selecting Courses --- 二分图最大匹配

匈牙利算法模板题 有n门课程,每门课程可能有不同一时候间,不同一时候间的课程等价. 问不冲突的情况下最多能选多少门课. 建立二分图,一边顶点表示不同课程,还有一边表示课程的时间(hash一下). #include <iostream> #include <cstring> #include <string> #include <cstdio> #include <cmath> #include <algorithm> #include

POJ 1469 COURSES 二分图最大匹配

就是判断一下是不是每一个课程都能找到自己的代表人,做一遍最大匹配看看匹配数是否等于p即可 #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #include <climits> #include <string> #include <iostream> #include <map> #include <cs

HDU ACM 1083 Courses 二分图最大匹配

题意:p门课,每门课有若干学生,要为每门课分配一名课代表,每个学生只能担任一门课的课代表,若每个课都能找到课代表,则输出"YES",否则"NO". 分析:二分图的最大匹配,对课程.学生关系建立一个图,进行二分图最大匹配,当最大匹配数==课程数时说明能够满足要求,否则不能. #include<iostream> using namespace std; #define N 303 bool cs[N][N]; //cs[i][j]表示学生j是否选i这个课程

POJ 1469 COURSES (二分图最大匹配 匈牙利算法)

COURSES Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 18892   Accepted: 7455 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 poss

[POJ] 2239 Selecting Courses(二分图最大匹配)

题目地址:http://poj.org/problem?id=2239 Li Ming大学选课,每天12节课,每周7天,每种同样的课可能有多节分布在不同天的不同节.问Li Ming最多可以选多少节课.把n种课划分为X集合,把一周的84节课划分为Y集合, 从Xi向Yi连边,那么就转化成了求二分图的最大匹配数,然后匈牙利算法就可以了. 1 #include<cstdio> 2 #include<iostream> 3 #include<string.h> 4 #includ

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 Limit: 65536/32768 K (Java/Others)Total Submission(s): 8869    Accepted Submission(s): 4319 Problem Description Consider a group of N students and P c

Poj 1469 COURSES -二分图最大匹配裸题

题目:问course是否有完美匹配 /************************************************ Author :DarkTong Created Time :2016/7/30 22:28:35 File Name :Poj_1469.cpp *************************************************/ #include <cstdio> #include <cstring> #include <

POJ 1469 COURSES【二分图最大匹配】

分析:二分图最大匹配 代码: 1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <vector> 5 using namespace std; 6 7 const int maxn = 305; 8 9 int n; 10 11 vector<int> G[maxn]; 12 int vis[maxn]; 13 int Link[maxn]; 14

POJ1469 COURSES 【二分图最大匹配&amp;#183;HK算法】

COURSES Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 17777   Accepted: 7007 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 poss