poj1469 COURSES(匈牙利算法)(解题报告)
来源:互联网 发布:淘宝定时上架 编辑:程序博客网 时间:2024/05/16 01:31
链接:
http://poj.org/problem?id=1469
COURSES
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.
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
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
思路:匹配问题,匈牙利算法的的模板套用
#include<stdio.h>#include<string.h>#include<iostream>using namespace std;const int m1=310,m2=110;int result[m1],state[m1],data[m2][m1];int n,p;int find(int a){for(int i=1;i<=n;i++){if(data[a][i]==1&&!state[i]){ state[i]=1; if(result[i]==0||find(result[i])) {result[i]=a; return 1; } }}return 0;} int main(){ int i,j,ans,t; int c,num; while(~scanf("%d",&t)) { while(t--) { ans=0; memset(data,0,sizeof(data)); memset(result,0,sizeof(result)); scanf("%d%d",&p,&n); for(i=1;i<=p;i++) { scanf("%d",&c); for(j=1;j<=c;j++) {scanf("%d",&num); data[i][num]=1; } } for(i=1;i<=p;i++) { memset(state,0,sizeof(state)); if(find(i))ans++; } if(ans==p) printf("YES\n"); else printf("NO\n"); } }return 0;}
- poj1469 COURSES(匈牙利算法)(解题报告)
- pku 匈牙利算法 1469 COURSES 解题报告
- poj1469匈牙利算法(模板)
- poj1469 COURSES(二分图)
- 【二分图最大匹配】【匈牙利算法】poj1469 COURSES && poj2446 Chessboard
- POJ1469[COURSES] 二分图最大匹配 匈牙利算法
- 二分图匹配匈牙利算法(poj1469验证)
- HDU1083+POJ1469 (匈牙利算法+最大二分匹配)
- HDU 1083 Courses( 匈牙利算法 )
- hdu1083——Courses(匈牙利算法)
- HDU-1083 Courses(匈牙利算法)
- poj1469--COURSES(二分图最大匹配)
- COURSES(poj1469,二分图最大匹配)
- hdu2063 过山车&&poj1469 COURSES(二分匹配)
- 匈牙利算法模版(poj1469)
- POJ 2239 Selecting Courses(二分图匹配——匈牙利算法||网络流)解题报告
- 解题报告-HDOJ-2063(最大二分匹配-匈牙利算法)
- 二分图,匈牙利算法,poj1469
- 第三篇博文?
- poj 1142Smith Numbers(解题报告)
- uva 10487 - Closest Sums
- Detour注意点及原理
- flash代码
- poj1469 COURSES(匈牙利算法)(解题报告)
- 最适合自己的类型简历
- ObjectDataSourse动态绑定数据及高效分页
- innodb 行锁实现算法
- hdu 3336 Count the string
- 如果令 A B C D E F G H I J K L M N O P Q R S T U V W X Y Z 分别等于
- 委托
- asinh 函数
- S5PV210 Android2.3 添加自定义按键:作唤醒功能