二分圖(bfs+匈牙利演算法) POJ 1469 course

COURSES
 

  

   

    
Time Limit: 1000MS
    
 
    
Memory Limit: 10000K
   
   

    
Total Submissions: 8076
    
 
    
Accepted: 3208
   
  
 
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
23 33 1 2 32 1 21 13 32 1 32 1 31 1
Sample Output
YESNO
/*題目大意：將vy個工作分配給vx個學生，求能不能每個同學都有分配到工作。題目解答：典型的二分圖，只要比較最大匹配數目等不等與學生數九可以了.代碼技巧：map[x][y]學生x能否勝任工作y，可以為1，不可以為0.match[y]=x,X集合中和工作y匹配的的學生編號.*/Source CodeProblem: 1469   User: wawadimu Memory: 204K   Time: 438MS Language: C++   Result: Accepted Source Code #include<iostream>#include<string>using namespace std;//記錄與mathch[y]匹配的x數值int match[305];bool vis[305];//map[i][j]==1表示元素i到元素j有一條邊bool map[105][305];//x集合和y集合的頂點數int vx,vy;//深度優先遍曆,從元素p出發尋找增廣路bool dfs(int x){    int t,y;    for(y=1;y<=vy;y++)    {//        if(vis[vx]) continue;        if(!vis[y] && map[x][y])        {            vis[y]=true;//            t=match[y];//            match[y]=x;//            if(t==0 || dfs(t)) return true;//            match[y]=t;//增廣路不成功,還原成原來匹配            if(match[y]==0 || dfs(match[y]))            {                match[y]=x;                return true;            }        }    }    return false;}int main(){    //freopen("1469.txt","r",stdin);    int i,j,k,n,Case,ans,cnt;    int m;    scanf("%d",&Case);    while(Case--)    {        scanf("%d%d",&vx,&vy);        memset(map,0,sizeof(map));        for(i=1;i<=vx;i++)        {            scanf("%d",&cnt);            for(j=1;j<=cnt;j++)            {                scanf("%d",&k);                map[i][k]=1;            }        }        ans=0;        memset(match,0,sizeof(match));        for(i=1;i<=vx;i++)        {            memset(vis,0,sizeof(vis));            //深度優先搜尋尋找增廣路,找到一條增廣路，匹配+1            if(dfs(i))ans++;//            for(int l=1;l<=vy;l++)//            printf("%d   ",match[l]);        }        if(ans==vx)        printf("YES/n");        else        printf("NO/n");    }}