【二分图】poj 1469 COURSES

COURSES

Time Limit: 1000MS
Memory Limit: 10000KTotal Submissions: 23715
Accepted: 9265

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
  

Input

Your 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 Student_{1 1} Student_{1 2} ... Student_{1 Count1}
Count2 Student_{2 1} Student_{2 2} ... Student_{2 Count2}
...
CountP Student_{P 1} Student_{P 2} ... Student_{P 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

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.

Sample Input

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

Sample Output

YESNO
题目大意：
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
///AC代码
#include <algorithm>#include <cmath>#include <cstdio>#include <cstring>#include <ctime>#include <iostream>#include <map>#include <queue>#include <set>#include <stack>#include <string>#include <vector>#define eps 1e-8#define INF 0x7fffffff#define maxn 100005#define PI acos(-1.0)using namespace std;typedef long long LL;const int N = 302;/*变种1：二分图的最小顶点覆盖：假如选了一个点就相当于覆盖了以它为端点的所有边，你需要选择最少的点来覆盖所有的边二分图的最小顶点覆盖数 = 二分图的最大匹配数变种2：DAG图（无回路有向图）的最小路径覆盖路径覆盖就是在图中找一些路经，使之覆盖了图中的所有顶点，且任何一个顶点有且只有一条路径与之关联，如果把这些路径中的每条路径从它的起始点走到它的终点，那么恰好可以经过图中的每个顶点一次且仅一次DAG图的最小路径覆盖数 = 节点数（n）- 最大匹配数（m）变种3: 二分图的最大独立集：在图中选取最多的点，使任意所选两点均不相连二分图的最大独立集数 = 节点数（n）- 最大匹配数（m）*//*=***************************************************二分图匹配（匈牙利算法的DFS实现）INIT：g[][]两边定点划分的情况CALL:res=hungary();输出最大匹配数优点：适于稠密图，DFS找增广路快，实现简洁易于理解时间复杂度:O(VE);***************************************************=*/const int MAXN = 1000;int uN, vN; //u,v数目int g[MAXN][MAXN];//编号是0~n-1的int linker[MAXN];bool used[MAXN];bool dfs(int u){    for (int v = 0; v < vN; v++)        if (g[u][v] && !used[v])        {            used[v] = true;            if (linker[v] == -1 || dfs(linker[v]))            {                linker[v] = u;                return true;            }        }    return false;}int hungary(){    int res = 0;    memset(linker, -1, sizeof(linker));    for (int u = 0; u < uN; u++)    {        memset(used, 0, sizeof(used));        if (dfs(u))        {            res++;        }    }    return res;}int main(){    int t;    scanf("%d", &t);    while (t--)    {        memset(g, 0, sizeof(g));        scanf("%d %d", &uN, &vN);        int ct, xx;        for (int i = 0; i < uN; i++)        {            scanf("%d", &ct);            while (ct--)            {                scanf("%d", &xx);                g[i][xx - 1] = 1;            }        }        int ans = hungary();        if (ans == uN)        {            printf("YES\n");        }        else        {            printf("NO\n");        }    }    return 0;}