二分图(bfs+匈牙利算法) POJ 1469 course

课程分配问题解析

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本文探讨了一个典型二分图问题——如何将多个工作合理地分配给若干名学生，确保每名学生都能获得一个任务，且每个任务仅由一名学生承担。通过深度优先搜索实现匈牙利算法来找出最大匹配，进而解决该问题。

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 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
/*
题目大意：将vy个工作分配给vx个学生，求能不能每个同学都有分配到工作。
题目解答：典型的二分图，只要比较最大匹配数目等不等与学生数九可以了.
代码技巧：map[x][y]学生x能否胜任工作y，可以为1，不可以为0.match[y]=x,X集合中和工作y匹配的的学生编号.
*/

Source Code

Problem: 1469   User: wawadimu 
Memory: 204K   Time: 438MS 
Language: C++   Result: Accepted 


Source Code 
#include<iostream>
#include<string>
using namespace std;
//记录与mathch[y]匹配的x数值
int match[305];
bool vis[305];
//map[i][j]==1表示元素i到元素j有一条边
bool map[105][305];
//x集合和y集合的顶点数
int vx,vy;
//深度优先遍历,从元素p出发寻找增广路
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;
//增广路不成功,还原成原来匹配
            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));
            //深度优先搜索寻找增广路,找到一条增广路，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");
    }
}