二分匹配最大匹配 PKU1469 - c++编程基础

一个学生可以有多种选择，问能否每个学生刚好选一门课，但是每门课最多只有一个学生可以选择

典型的二分匹配最大匹配，直接套模板

COURSES

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 17878		Accepted: 7048

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?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

Source

Southeastern Europe 2000

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                   using namespace std; bool g[110][310];//邻接矩阵，true代表有边相连 bool flag,vis[310];//记录v2中的某个点是否被搜索过 int match[310];//记录与v2中的点匹配的点的编号 int p,n;//二分图中左边，右边集合顶点的个数 //匈牙利算法 bool DFS(int u){ int i; for(i=1;i<=n;++i){ if(g[u][i] && !vis[i])//如果节点i与u相连并且没有被查找过 { vis[i] = true; if(match[i]==-1 || DFS(match[i]))//如果i没有在前一个匹配M中，或者i在匹配M中，但是从与i相邻的节点出发可以找到增广路径 { match[i] = u;//记录查找成功记录，更新匹配M（就是取反） return true;//返回查找成功 } } } return false; } int main(){ int i,j,k,t,v,ans; scanf("%d",&t); while(t--){ scanf("%d%d",&p,&n); for(i=1;i<=p;i++){ for(j=1;j<=n;j++){ g[i][j] = false; } } for(i=1;i<=n;i++){ match[i] = -1; } flag = true; for(i=1;i<=p;i++){ scanf("%d",&k); if(k == 0){ flag = false; } while(k--){ scanf("%d",&v); g[i][v] = true; } } if(flag==true){ ans = 0; for(i=1;i<=p;i++){ memset(vis,false,sizeof(vis));//清空上次搜索时的标记 if(DFS(i)==true){ ans++; } } if(ans == p){ printf("YES\n"); } else{ printf("NO\n"); } } else{ printf("NO\n"); } } return 0; }