COURSES POJ1469(模板)
阿新 • • 發佈:2018-11-07
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
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
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
Source
Southeastern Europe 2000
題解:
標準的匈牙利演算法,最需要最後的匹配樹等於課程數,我感覺最大流也是可以寫的。
#include <cstring> #include <cstdio> using namespace std; const int MAXN=50005; struct node{ int v,next; }edge[MAXN]; int cnt=0,head[MAXN]; int p,n; void add(int x,int y) { edge[cnt].v=y; edge[cnt].next=head[x]; head[x]=cnt++; } int ans=0; bool vis[MAXN]; int match[MAXN]; bool findpath(int x) { for (int i = head[x]; i !=-1 ; i=edge[i].next) { int v=edge[i].v; if(!vis[v]) { vis[v]=true; if(match[v]==-1||findpath(match[v])) { match[v]=x; return true; } } } return false; } void hungry() { for (int i = 1; i <=p; ++i) { memset(vis,false, sizeof(vis));//沒次都去初始化 if(findpath(i))//尋找是否有增廣路 ans++; } } int main() { int _; scanf("%d",&_); while(_--) { cnt=0; ans=0; memset(head,-1, sizeof(head)); memset(match,-1, sizeof(match)); memset(vis,false, sizeof(vis)); scanf("%d%d",&p,&n); int num,x; for (int i = 1; i <=p ; ++i) { scanf("%d",&num); while(num--) { scanf("%d",&x); add(i,x); } } hungry(); if(ans==p) printf("YES\n"); else puts("NO"); } return 0; }