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最新推荐文章于 2020-04-12 21:40:00 发布

san.hang

最新推荐文章于 2020-04-12 21:40:00 发布

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原文链接：http://www.cnblogs.com/Deng1185246160/p/3236356.html

本文介绍了一个关于课程代表分配的问题，即如何从多个学生中选择合适的代表来对应不同的课程，确保每位代表只代表一门课程且每门课程都有代表。文章提供了两种解决方案，一种使用前向星图论技巧，另一种则是直接实现的二分图匹配算法。

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COURSES

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 15394		Accepted: 6082

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

Sample Output

YES
NO

/*poj 1469课程 请用scanf
    数据范围好像有错
*/
#include<iostream>
using namespace std;
const int N = 305; //学生数
const int P = 105; //课程数   
int NUM;           //前向星变量
int head[N];       //前向星表头
int match_x[N];    //学生有没有被匹配
int match_y[N];    //课程有没有被匹配
bool mark_y[N];    //课程在找增广路时有没有被访问过

struct Node{
    int u,v,next; 
}node[N*P]; //N个学生，没个P门课

void add(int a,int b){ //前向星添加边
    node[NUM].u = a;
    node[NUM].v = b;
    node[NUM].next = head[a];
    head[a] = NUM++;
}

bool path(int u){
    for(int i=head[u]; ~i ;i=node[i].next){  //枚举每个能和u匹配的点
        int v = node[i].v;                   //匹配的点v
        if(mark_y[v] == false){              //如果本轮搜索没被标记过
            mark_y[v] = true;                //标记为搜索过
            if(match_y[v]==-1 || path(match_y[v])){//如果没有匹配 或 找到增广路
                match_x[u] = v;              //点u标记为v匹配
                match_y[v] = u;              //点v标记为跟u
                return true;                 //返回1
            }
        }
    }
    return false;                            //匹配不到返回0
}

int slove(int n){
    memset(match_x,255,sizeof(match_x));    //初始化，-1表示没有匹配到
    memset(match_y,255,sizeof(match_y));
    int result = 0;                         //答案 
    for(int i=1;i<=n;i++){                  //枚举没个X集合的点（学生）
        if(match_x[i] == -1){               //如果该学生没匹配过
            memset(mark_y,false,sizeof(mark_y)); //进行匹配
            result += path(i);              //结果加匹配结果
        }
    }
    return result;                         //返回输出
}

int main(){
    int t,n,p,m,to;
    while(scanf("%d",&t)!=EOF){
        while(t--){
            memset(head,255,sizeof(head));    //初始化连表头
            NUM = 0;                          //初始化边数
            scanf("%d%d",&p,&n);
            for(int i=1;i<=p;i++){            //第i个学生跟m门课程匹配
                scanf("%d",&m);
                while(m--){                   //通过前向星添加变
                    scanf("%d",&to);
                    add(i,to);
                }
            }
            if(p==slove(n)) printf("YES\n"); //若每个学生都能做课代表（完美匹配）
            else printf("NO\n");             
        }
    }
    return 0;
}

不用前向星做：

#include <cstdio>
#include <cstring>
#define M 301

int P, N;   //课程数; 学生数
bool course[M][M], used[M]; 
int match[M]; 

bool DFS( int k )    //DFS增广
{
    int i, temp; 
    for( i = 1; i <= N; i++ )
    {
        if( course[k][i] && !used[i] )
        {
            used[i] = 1; 
            temp = match[i];  match[i] = k; 
            if( temp == -1 || DFS( temp ) ) 
                return 1; 
            match[i] = temp; 
        }
    }
    return 0; 
}

int MaxMatch()    //求二部图最大匹配
{
    int i, MatchNum = 0; 
    memset( match, -1, sizeof( match ) ); 
    for( i = 1; i <= P; i++ )
    {
        memset( used, 0, sizeof( used ) ); 
        if( DFS( i ) ) MatchNum++; //累加匹配数
        if( MatchNum == P ) break; 
    }
    return MatchNum;    //返回最大匹配数
}

int main( )
{
    int i, j, num, t, n; 
    scanf( "%d", &n ); 
    while( n-- )
    {
        scanf( "%d%d", &P, &N );    //读入数据
        memset( course, 0, sizeof( course ) ); 
        for( i = 1; i <= P; i++ )
        {
            scanf( "%d", &num ); 
            for( j = 0; j < num; j++ )
            {
                scanf( "%d", &t ); 
                course[i][t] = 1; 
            }
        }
        //如果匹配数与课程数相等,则可以组成委员会
        if( MaxMatch() == P ) printf( "YES\n" ); 
        else printf( "NO\n" ); 
    }
    return 0; 
}

转载于:https://www.cnblogs.com/Deng1185246160/p/3236356.html