COURSES
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
题目类型：二分图最大匹配
算法分析：只要二分图的最大匹配能够将课程都“覆盖”住，则输出“YES”，反之输出“NO”
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#include <set> #include <bitset> #include <list> #include <map> #include <stack> #include <queue> #include <deque> #include <string> #include <vector> #include <ios> #include <iostream> #include <fstream> #include <sstream> #include <iomanip> #include <algorithm> #include <utility> #include <complex> #include <numeric> #include <functional> #include <cmath> #include <ctime> #include <climits> #include <cstdarg> #include <cstdio> #include <cstdlib> #include <cstring> #include <cctype> #include <cassert> using namespace std; #define CFF freopen ("aaa.txt", "r", stdin) #define CPPFF ifstream cin ("aaa.txt") #define i64 long long const int INF = 0x7FFFFFFF; const int MOD = 1e9 + 7; const double EPS = 1e-10; const double PI = 2 * acos (0.0); const int maxn = 300 + 66; int edge[maxn][maxn], vis[maxn]; int cx[maxn], cy[maxn], cxlen, cylen; void Init () { for (int i = 1; i < maxn; i++) for (int j = 1; j < maxn; j++) edge[i][j] = INF; } long long dfs (int u) { for (int v = 1; v <= cylen; v++) { if (edge[u][v] < INF && !vis[v]) { vis[v] = 1; if (cy[v] == -1 || dfs (cy[v])) { cy[v] = u; cx[u] = v; return 1LL; } } } return 0LL; } long long MaxMatch () { long long res = 0; memset (cx, -1, sizeof (cx)); memset (cy, -1, sizeof (cy)); for (int i = 1; i <= cxlen; i++) if (cx[i] == -1) { memset (vis, 0, sizeof (vis)); res += dfs (i); } return res; } int main() { // CFF; int t; scanf ("%d", &t); while (t--) { Init(); scanf ("%d%d", &cxlen, &cylen); for (int i = 1; i <= cxlen; i++) { int tt; scanf ("%d", &tt); for (int j = 1; j <= tt; j++) { int a; scanf ("%d", &a); edge[i][a] = 1; } } long long res = MaxMatch(); if (res == cxlen) puts ("YES"); else puts ("NO"); } return 0; } |
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