HDU1083（二分图最大匹配vector实现）

Courses

Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5153    Accepted Submission(s): 2477

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

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

 

Source

Southeastern Europe 2000

 

题意：有p个课程和n个学生，每个学生可以自由选择课程（0到p个），现在要建立一个委员会，问是否能找到每个课程都有学生代表的集合，一个学生只能代表一个课程

简单的二分匹配，主要是尝试用vector实现一下

/*
ID: LinKArftc
PROG: 1083.cpp
LANG: C++
*/

#include <map>
#include <set>
#include <cmath>
#include <stack>
#include <queue>
#include <vector>
#include <cstdio>
#include <string>
#include <utility>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define eps 1e-8
#define randin srand((unsigned int)time(NULL))
#define input freopen("input.txt","r",stdin)
#define debug(s) cout << "s = " << s << endl;
#define outstars cout << "*************" << endl;
const double PI = acos(-1.0);
const double e = exp(1.0);
const int inf = 0x3f3f3f3f;
const int INF = 0x7fffffff;
typedef long long ll;

const int maxn = 310;
int uN, vN;
vector <int> vec[maxn];
int linker[maxn];
bool vis[maxn];

bool dfs(int u) {
    int cnt = vec[u].size();
    for (int i = 0; i < cnt; i ++) {
        int v = vec[u][i];
        if (!vis[v]) {
            vis[v] = true;
            if (linker[v] == -1 || dfs(linker[v])) {
                linker[v] = u;
                return true;
            }
        }
    }
    return false;
}

int hungry() {
    memset(linker, -1, sizeof(linker));
    int ret = 0;
    for (int i = 1; i <= uN; i ++) {
        memset(vis, 0, sizeof(vis));
        if (dfs(i)) ret ++;
    }
    return ret;
}

int main() {
    int T, cnt, u, v;
    scanf("%d", &T);
    while (T -- ) {
        scanf("%d %d", &uN, &vN);
        for (int i = 1; i <= uN; i ++) vec[i].clear();
        for (int i = 1; i <= uN; i ++) {
            scanf("%d", &cnt);
            for (int j = 1; j <= cnt; j ++) {
                scanf("%d", &v);
                vec[i].push_back(v);
            }
        }
        if (uN == hungry()) printf("YES
");
        else printf("NO
");
    }

    return 0;
}