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2013
11-28

HDU 1083 Courses-分治-[解题报告] C++

Courses

问题描述 :

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:

样例输入:

2
3 3
3 1 2 3
2 1 2
1 1
3 3
2 1 3
2 1 3
1 1

样例输出:

YES
NO 

给定一个二分图G,M为G边集的一个子集,如果M满足当中的任意两条边都不依附于同一个顶点,则称M是一个匹配

最大匹配即最大的匹配个数。

关于二分匹配的基本问题:http://www.cnblogs.com/heat-man/archive/2013/03/26/2982644.html

hdu 1083   courses(http://acm.hdu.edu.cn/showproblem.php?pid=1083

本题题目叙述有点乱,英语不好看了很久,有p门课,n个学生,每个学生可以选修1-p门课,给出了每一门课的学生选择信息(学生的序号),现在要成一个委员会,需满足每一门课都有一个选修了该课程的学生来做代表,每个学生只能代表一门课程,问能否组成委员会,如果能,则输出“YES”,否则“NO”;

二分图的最大匹配问题,以课程-学生建立二分图,求出最大匹配数等于课程数即满足。

#include<iostream>
 #include<cstring>
 #include<cstdio>
 using namespace std;
 #define N 305
 int match[N][N];
 int mark[N];
 int visted[N];
 int n,p;
 int t ;
 int DFS(int x)
 {
     int j;
     for(j=1;j<=n;j++)
     {
         if(!visted[j]&&match[x][j]==1)
         {
             visted[j]=1;
             if(!mark[j]||DFS(mark[j]))
             {
                 mark[j]=x;
                 return 1;
             }
         }
     }
     return 0;
 }
 int main()
 {
     //freopen("input.txt","r",stdin);
       // freopen("output.txt","w",stdout);
     scanf("%d",&t);
     while(t--)
     {
 
         memset(match,0,sizeof(match));
         memset(mark,0,sizeof(mark));
         scanf("%d%d",&p,&n);
         if(p>n)
         {
             printf("NO\n");
             continue ;
         }
         int i;
         for(i=1;i<=p;i++)
         {
             int num;
             int j;
             scanf("%d",&num);
             for(j=1;j<=num;j++)
             {
                 int a;
                 scanf("%d",&a);
                 match[i][a]=1;
 
             }
         }
         int  ans=0;
         for(i=1;i<=p;i++)
         {
             memset(visted,0,sizeof(visted));
             if(DFS(i))
                 ans++;
         }
         if(ans==p)
             printf("YES\n");
         else
             printf("NO\n");
     }
     return 0;
 }

 


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  2. 有限自动机在ACM中是必须掌握的算法,实际上在面试当中几乎不可能让你单独的去实现这个算法,如果有题目要用到有限自动机来降低时间复杂度,那么这种面试题应该属于很难的级别了。

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