poj——1469 COURSES
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 24192 | Accepted: 9426 |
Description
- 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
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
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:一些课一些人,组成一个集体,集体中每人代表每门不同的课,每门课在集体中有一名成员,问是否能组成这样的集体
思路:
这个题我们只需要求出二分图的最大匹配数,这样我们求出他们选的课程数,然后判断最大匹配数是否等于课程总数。
代码:
#include<cstdio> #include<cstdlib> #include<cstring> #include<iostream> #include<algorithm> #define N 501 using namespace std; bool vis[N]; int t,w,n,m,x,ans,girl[N],map[N][N]; int read() { ,f=; char ch=getchar(); ; ch=getchar();} +ch-';ch=getchar();} return x*f; } int find(int x) { ;i<=m;i++) { if(!vis[i]&&map[x][i]) { vis[i]=true; ||find(girl[i])) {girl[i]=x;;} } } ; } int main() { t=read(); while(t--) { ans=; memset(map,,sizeof(map)); n=read(),m=read(); ;i<=n;i++) { w=read(); ;} } if(m<n){printf("NO\n"); continue;} memset(girl,-,sizeof(girl)); ;i<=n;i++) { memset(vis,,sizeof(vis)); if(find(i)) ans++; } if(ans==n) printf("YES\n"); else printf("NO\n"); } ; }
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