POJ2239_Selecting Courses (maximum matching of the Bipartite Graph), selectingcourses

POJ2239_Selecting Courses (maximum matching of the Bipartite Graph), selectingcourses

Solution report

Http://blog.csdn.net/juncoder/article/details/38154699

Question Portal

Question:

There are 12 classes every day, 7 days a week, and one class has more than one day a week. Ask for a maximum of a few lessons in a week.

Ideas:

Take the course as a collection, and the course time as a collection. The bipartite graph is displayed.

#include <cstdio>#include <cstring>#include <iostream>using namespace std;int n,day[10][15],mmap[500][500],vis[500],cnt,pre[500];int dfs(int x){    for(int i=n+1;i<cnt;i++){        if(!vis[i]&&mmap[x][i]){            vis[i]=1;            if(pre[i]==-1||dfs(pre[i])){                pre[i]=x;                return 1;            }        }    }    return 0;}int main(){    int i,j,a,b,t;    while(cin>>n){        cnt=n+1;        memset(mmap,0,sizeof(mmap));        memset(day,0,sizeof(day));        memset(pre,-1,sizeof(pre));        for(i=1;i<=n;i++){            cin>>t;            for(j=1;j<=t;j++){                cin>>a>>b;                if(!day[a][b])                {                    day[a][b]=cnt++;                }                mmap[i][day[a][b]]=1;            }        }        int ans=0;        for(i=1;i<=n;i++){            memset(vis,0,sizeof(vis));            ans+=dfs(i);        }        cout<<ans<<endl;    }    return 0;}

Selecting Courses

Time Limit:1000 MS		Memory Limit:65536 K
Total Submissions:8466		Accepted:3769

Description

It is well known that it is not easy to select courses in the college, for there is usually conflict among the time of the courses. li Ming is a student who loves study every much, and at the beginning of each term, he always wants to select courses as more as possible. of course there shocould be no conflict among the courses he selects.

There are 12 classes every day, and 7 days every week. there are hundreds of courses in the college, and teaching a course needs one class each week. to give students more convenience, though teaching a course needs only one class, a course will be taught several times in a week. for example, a course may be taught both at the 7-th class on Tuesday and 12-th class on Wednesday, you showould assume th At there is no difference between the two classes, and that students can select any class to go. at the different weeks, a student can even go to different class as his wish. because there are so easy courses in the college, selecting courses is not an easy job for Li Ming. as his good friends, can you help him?

Input

The input contains several cases. for each case, the first line contains an integer n (1 <= n <= 300), the number of courses in Li Ming's college. the following n lines represent n different courses. in each line, the first number is an integer t (1 <= t <= 7*12), the different time when students can go to study the course. then come t pairs of integers p (1 <= p <= 7) and q (1 <= q <= 12 ), which mean that the course will be taught at the q-th class on the p-th day of a week.

Output

For each test case, output one integer, which is the maximum number of courses Li Ming can select.

Sample Input

51 1 12 1 1 2 21 2 22 3 2 3 31 3 3

Sample Output

4

What is the matching of a bipartite graph? maximum matching: maximum weighted matching

Given a bipartite graph G, In a subgraph M of G, any two edges in the edge set of M are not attached to the same vertex. M is a matching.
Selecting the largest subset of the number of edges is called the maximal matching problem)
If in a match, each vertex in the graph is associated with an edge in the graph, it is called a full match or a complete match.
The maximum stream or Hungary algorithm can be used to obtain the maximum matching of a bipartite graph.
Reference: bk.baidu.com/view/503667.htm

Matlab program for maximum matching of bipartite graphs

[Num h] = maxnum (g); % g is the bipartite graph ing matrix %. A self-written maxnum function is called. The return value num is the maximum value, and h is the hij (not unique) maxnum. m content, which uses the Hungary algorithm also uses a recursive incpath function to find the augmented path function [num h] = maxnum (g) s = size (g ); global G_h; % matrix hij record selected global G_g; % matrix gij record matched global G_v; % record current path accessed node G_h = false (s ); % matrix hij is initially blank G_g = g; % matrix gij is the passed parameter gfor I = 1: s (1) G_v = false (1, s (2 )); % The Node accessed by each initialization path is empty incpath (I); % search for the augmented path endh = G_h from Ai; num = sum (h (:)); % The maximum number of matches output, and the matching matrix hclear global 'G _ H'; clear Global 'G _ G'; endfunction OK = incpath (I) % starting from Ai global G_h; global G_g; global G_v; OK = false; j = find (~ G_h (I, :) & G_g (I ,:)&~ G_v, 1); % find the condition Bjif isempty (j), return; end % cannot find return falseG_v (j) = true; % found, mark Bj as Access Node ii = find (G_h (:, j); % find Bj in original match if isempty (ii) % if not in original match G_h (I, j) = true; OK = true; return; end % find the end of the augmented path and return trueok = incpath (ii); % if the original match is in progress, call incpath recursively to find if OK % based on the matching Aii. if the recursive search returns G_h (I, j) = ~ G_h (I, j); G_h (ii, j) = ~ G_h (ii, j); OK = true; return; end % PATH Reversed return trueend