POJ 1469 COURSES //简单二分图
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COURSES
Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 17211 Accepted: 6769
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.
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
Source
Southeastern Europe 2000
#include <stdio.h>#include <string.h>bool map[105][305];//图,记录 第i个同学 选修 第j个课程bool visit[305];//记录该点是否被访问过int r_jilu[305];//记录右边项 对应的 左边点bool dfs(int n, int r);int main(){int t;int l, r;int ans, k, v;int i;scanf("%d", &t);while(t--){scanf("%d%d", &l, &r);memset(map, false, sizeof(map));for(i=1; i<=r; i++)r_jilu[i] = -1;//初始化for(i=1; i<=l; i++){scanf("%d", &k);while(k--){scanf("%d", &v);map[i][v] = true;}}ans = 0;for(i=1; i<=l; i++){memset(visit, false, sizeof(visit));if( dfs(i, r) )ans++;}if(ans == l)printf("YES\n");elseprintf("NO\n");}return 0;}bool dfs(int n, int r)// 匈牙利算法 {int i;for(i=1; i<=r; i++){if(map[n][i]==true && visit[i]==false)//当此点为左边对应的点,并且没被访问过时{visit[i] = true;if(r_jilu[i]==-1 || dfs(r_jilu[i], r))//如果i未在前一个匹配M中,或者i在匹配M中,但是从与i相邻的节点出发可以有增广路径{r_jilu[i] = n;//更新return true;}}}return false;}
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