描述

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

輸入

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?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.

輸出

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.

樣例輸入

2
3 3
3 1 2 3
2 1 2
1 1
3 3
2 1 3
2 1 3
1 1

樣例輸出

YES
NO

題意

有n個學生和p門課，每門課有對應的學生要求，判斷能否選出p個學生剛好上p門課

題解

一道很裸的二分圖匹配題，剛好拿來熟悉下算法

這里用匈牙利算法，判斷p門課程是否都能成功匹配

代碼

1 #include<bits/stdc++.h> 2 using namespace std; 3 4 const int N=305,P=105; 5 int G[P][N],vis[N],match[P]; 6 int n,p; 7 int Find(int u) 8 { 9 for(int i=1;i<=n;i++) 10 { 11 if(G[u][i]&&!vis[i]) 12 { 13 vis[i]=1; 14 if(!match[i]||Find(match[i])) 15 { 16 match[i]=u; 17 return 1; 18 } 19 } 20 } 21 return 0; 22 } 23 int main() 24 { 25 int k,t,stu; 26 cin>>t; 27 while(t--) 28 { 29 memset(G,0,sizeof(G)); 30 cin>>p>>n; 31 for(int i=1;i<=p;i++) 32 { 33 cin>>k; 34 for(int j=0;j<k;j++) 35 { 36 cin>>stu; 37 G[i][stu]=1; 38 } 39 } 40 int flag=1; 41 memset(match,0,sizeof(match)); 42 for(int i=1;i<=p;i++) 43 { 44 memset(vis,0,sizeof(vis)); 45 if(!Find(i)) 46 { 47 flag=0;break; 48 } 49 } 50 printf("%s\n",flag?"YES":"NO"); 51 } 52 }

轉載于:https://www.cnblogs.com/taozi1115402474/p/8724297.html

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