POJ 1469 COURSES(二部图匹配)

Posted 贱人方

tags:

篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了POJ 1469 COURSES(二部图匹配)相关的知识,希望对你有一定的参考价值。

                                                                 COURSES
Time Limit: 1000MS   Memory Limit: 10000K
Total Submissions: 21527   Accepted: 8460

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 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

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
【分析】最大匹配模板题。
#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <string>
#include <vector>
#define inf 0x7fffffff
#define met(a,b) memset(a,b,sizeof a)
typedef long long ll;
using namespace std;
const int N = 405;
const int M = 40005;
int n,m,cnt;
int x[M],y[M];
int t[M],head[N];
struct man
{
    int to,next;
}g[M];
bool dfs(int u) {
    for(int i=head[u];i!=-1;i=g[i].next) {
        int v=g[i].to;
        if(!t[v]) {
            t[v]=1;
            if(!y[v]||dfs(y[v])) {
                x[u]=v;
                y[v]=u;
                return true;
            }
        }
    }
    return false;
}
void MaxMatch() {
    int ans=0;
    for(int i=1; i<=m; i++) {
        if(!x[i]) {
            met(t,0);
            if(dfs(i))ans++;
        }
    }
    if(ans==m)printf("YES\\n");
    else printf("NO\\n");
}
void add(int u,int v) {
    g[cnt].to=v;g[cnt].next=head[u];head[u]=cnt++;
}
int main() {
    int T;
    int k,number;
    bool flag;
    scanf("%d",&T);
    for(int k=1; k<=T; k++) {
        met(g,0);
        met(x,0);
        met(y,0);
        met(head,-1);cnt=0;
        scanf("%d%d",&m,&n);
        for(int i=1; i<=m; i++) {
            int t,u;
            scanf("%d",&t);
            while(t--) {
                scanf("%d",&u);
                add(i,u);
            }
        }
        MaxMatch();
    }
    return 0;
}
View Code

以上是关于POJ 1469 COURSES(二部图匹配)的主要内容,如果未能解决你的问题,请参考以下文章

二分图匹配入门专题1F - COURSES poj1469最大匹配--匈牙利算法模板题

POJ1469 COURSES 二分图最大匹配&#183;HK算法

POJ1469 COURSES 二分图匹配 匈牙利算法

poj 1469 COURSES(二分匹配模板)

POJ 1469 -- COURSES (二分匹配)

POJ - 1469 COURSES [二分图匈牙利算法]