POJ 1469-COURSES（二分图匹配入门-匈牙利算法）

COURSES

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 21307 Accepted: 8374

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} ... Student_{1 Count1} 
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

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

题目意思：

N个学生，P门课，每个学生学习P门课中的0~P门。

是判断能否选出P个学生，组成一个委员会，满足：

①委员会中每名学生代表一门不同的课程（学习了就能代表这门课）；

②每门课在委员会中有一名代表。

解题思路：

二分图匹配入门大水题，直接用匈牙利算法切掉。注意用scanf，cin和cout会TLE…

P门课与N个学生之间以“是否学习”连边建图。

#include <iostream>#include <cstdio>#include <cstring>#include <vector>#include <queue>#include <algorithm>#define maxn 1010#define INF 0xfffffffusing namespace std;int nx,ny;//X与Y集合中元素的个数int g[maxn][maxn];//边的邻接矩阵int cx[maxn],cy[maxn];//cx[i]表示最终求得的最大匹配中与Xi匹配的顶点，cy[i]同理int mk[maxn];//mark，记录顶点访问状态int path(int u){    for(int v=1; v<=ny; ++v)        if(g[u][v]&&!mk[v])        {            mk[v]=1;            if(cy[v]==-1||path(cy[v]))            {                cx[u]=v;                cy[v]=u;                return 1;            }        }    return 0;}int MaxMatch(){    int res=0;    memset(cx,-1,sizeof(cx));    memset(cy,-1,sizeof(cy));    for(int i=1; i<=nx; ++i)        if(cx[i]==-1)        {            memset(mk,0,sizeof(mk));            res+=path(i);            if(res==nx) break;        }    return res;}int main(){    int t;    scanf("%d",&t);    while(t--)    {        scanf("%d%d",&nx,&ny);        memset(g,0,sizeof(g));        int num,i,j,x;        for(i=1; i<=nx; ++i)        {            scanf("%d",&num);            for(j=0; j<num; ++j)            {                scanf("%d",&x);                g[i][x]=1;            }        }        if(MaxMatch()==nx) puts("YES");        else puts("NO");    }    return 0;}/*23 33 1 2 32 1 21 13 32 1 32 1 31 1*/