POJ 1469-COURSES（二分圖匹配入門-匈牙利演算法）

kewlgrl發表於2016-08-17
演算法
COURSES
Time Limit: 1000MS   Memory Limit: 10000K
Total Submissions: 21307   Accepted: 8374

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

Source

Southeastern Europe 2000

題目意思：

N個學生，P門課，每個學生學習P門課中的0~P門。

是判斷能否選出P個學生，組成一個委員會，滿足：

①委員會中每名學生代表一門不同的課程（學習了就能代表這門課）；

②每門課在委員會中有一名代表。

解題思路：

二分圖匹配入門大水題，直接用匈牙利演算法切掉。注意用scanf，cin和cout會TLE…

P門課與N個學生之間以“是否學習”連邊建圖。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <queue>
#include <algorithm>
#define maxn 1010
#define INF 0xfffffff
using namespace std;
int nx,ny;//X與Y集合中元素的個數
int g[maxn][maxn];//邊的鄰接矩陣
int cx[maxn],cy[maxn];//cx[i]表示最終求得的最大匹配中與Xi匹配的頂點，cy[i]同理

int mk[maxn];//mark，記錄頂點訪問狀態
int path(int u)
{
    for(int v=1; v<=ny; ++v)
        if(g[u][v]&&!mk[v])
        {
            mk[v]=1;
            if(cy[v]==-1||path(cy[v]))
            {
                cx[u]=v;
                cy[v]=u;
                return 1;
            }
        }
    return 0;
}
int MaxMatch()
{
    int res=0;
    memset(cx,-1,sizeof(cx));
    memset(cy,-1,sizeof(cy));
    for(int i=1; i<=nx; ++i)
        if(cx[i]==-1)
        {
            memset(mk,0,sizeof(mk));
            res+=path(i);
            if(res==nx) break;
        }
    return res;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&nx,&ny);
        memset(g,0,sizeof(g));
        int num,i,j,x;
        for(i=1; i<=nx; ++i)
        {
            scanf("%d",&num);
            for(j=0; j<num; ++j)
            {
                scanf("%d",&x);
                g[i][x]=1;
            }
        }
        if(MaxMatch()==nx) puts("YES");
        else puts("NO");
    }
    return 0;
}
/*
2
3 3
3 1 2 3
2 1 2
1 1
3 3
2 1 3
2 1 3
1 1
*/


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