zoukankan      html  css  js  c++  java
  • POJ 1469 COURSES


    COURSES
    Time Limit: 1000MS   Memory Limit: 10000K
    Total Submissions: 20478   Accepted: 8056

    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


    匈牙利算法,二分图匹配

    基本算是模板题了


    #include<cstdio>
    #include<iostream>
    #include<cmath>
    #include<cstring>
    #include<algorithm>
    using namespace std;
    int map[420][420];//二分图 
    int v[420];
    int cx[420];
    int cnt;
    int n,p;
    int dfs(int x){
    	int i;
    	for(i=1;i<=n;i++){
    		if(map[x][i] && !v[i]){
    			v[i]=1;
    			int temp=cx[i];
    			cx[i]=x;
    			if(temp==-1 ||dfs(temp))return 1;
    			cx[i]=temp;
    		}
    	}
    	return 0;
    }
    int mxmh(){
    	memset(cx,-1,sizeof(cx));
    	int i,ans=0;
    	for(i=1;i<=p;i++){
    		memset(v,0,sizeof(v));
    		if(dfs(i))ans++;
    		if(ans==p)break;
    	}
    	return ans;
    }
    int main(){
    	int T;
    	scanf("%d",&T);
    	while(T--){
    		memset(map,0,sizeof(map));	
    		scanf("%d%d",&p,&n);
    		int i,j,num,x;
    		for(i=1;i<=p;i++){
    			scanf("%d",&num);
    			for(j=1;j<=num;j++){
    				scanf("%d",&x);
    				map[i][x]=1;
    			}
    		}
    		if(mxmh()==p)printf("YES
    ");
    		else printf("NO
    ");
    	}
    	return 0;
    }







  • 相关阅读:
    jquery 第二节 Dom和jQuery的互相转换
    jquery 第一节 什么是jQuery
    SQL四大语句、四大完整性、五大约束
    empty和is_null以及isset函数在0、”0”、‘空串’、NULL、false、array()的计算值
    WAMP常用环境配置
    解读Java内部类
    每日编程系列——暗黑的字符串
    每日编程系列——跳石板
    每日编程系列——优雅的点
    每日编程系列——回文序列
  • 原文地址:https://www.cnblogs.com/SilverNebula/p/5550582.html
Copyright © 2011-2022 走看看