HDU 1083 Courses（二分匹配匈牙利算法模板题啊）

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

Problem 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

23 33 1 2 32 1 21 13 32 1 32 1 31 1

 

Sample Output

YESNO 

 

Source

Southeastern Europe 2000

题意：

一共有 p 门课，每门课都有若干的学生，现在要为每个课程选一名课代表，

每个学生只能担任一门课的课代表，如果每个课都能找到课代表，则输出"YES"，否则"NO"。

PS：

二分图最大匹配，对课程—学生关系建立一个图，进行二分图的最大匹配，

如果最大匹配数==课程数，说明能够满足要求，否则不能。

代码如下：

#include<cstdio>#include<cstring>#include<algorithm>#include<iostream>using namespace std;/* **************************************************************************//二分图匹配（匈牙利算法的DFS实现）//初始化：g[][]两边顶点的划分情况//建立g[i][j]表示i->j的有向边就可以了，是左边向右边的匹配//g没有边相连则初始化为0//L是匹配左边的顶点数，R是匹配右边的顶点数//调用：res=hungary();输出最大匹配数//优点：适用于稠密图，DFS找增广路，实现简洁易于理解//时间复杂度:O(VE)//***************************************************************************///顶点编号从1开始的#define MAXN 317int LN,RN;//L,R数目int g[MAXN][MAXN], linker[MAXN];bool used[MAXN];int dfs(int L)//从左边开始找增广路径{    int R;    for(R = 1; R <= RN; R++)    {        if(g[L][R]!=0 && !used[R])        {            //找增广路，反向            used[R]=true;            if(linker[R] == -1 || dfs(linker[R]))            {                linker[R]=L;                return 1;            }        }    }    return 0;//这个不要忘了，经常忘记这句}int hungary(){    int res = 0 ;    int L;    memset(linker,-1,sizeof(linker));    for( L = 1; L <= LN; L++)    {        memset(used,0,sizeof(used));        if(dfs(L) != 0)            res++;    }    return res;}int main(){    int t;    int p, n;    int k, L, R;    scanf("%d",&t);    while(t--)    {        scanf("%d%d",&p,&n);        memset(g,0,sizeof(g));        for(int i = 1; i <= p; i++)        {            scanf("%d",&k);            for(int j = 1; j <= k; j++)            {                scanf("%d",&R);                L = i;                g[L][R] = 1;            }        }        LN = p;        RN = n;        int res = hungary();//最大匹配数        //printf("%d\n",res);        if(res == p)        {            printf("YES\n");        }        else        {            printf("NO\n");        }    }    return 0 ;}