Time Limit: 1000MS | Memory Limit: 10000K |
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
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
//邻接矩阵 #include <cstdio> #include <cstring> using namespace std; int p, n; const int maxp = 110; const int maxn = 310; bool g[maxp][maxn]; int linker[maxn]; bool vis[maxn]; void input() { scanf("%d%d", &p, &n); int t, v; memset(g, 0, sizeof(g)); for (int i = 1; i <= p; i++) { scanf("%d", &t); while (t--) { scanf("%d", &v); g[i][v] = 1; } } } bool dfs(int u) { for (int v = 1; v <= n; v++) if (g[u][v] && !vis[v]) { vis[v] = 1; if (linker[v] == -1 || dfs(linker[v])) { linker[v] = u; return 1; } } return 0; } int hungary() { int res = 0; memset(linker, -1, sizeof(linker)); for (int u = 1; u <= p; u++) { memset(vis, 0, sizeof(vis)); if (dfs(u)) res++; } return res; } void solve() { if (hungary() == p) printf("YES\n"); else printf("NO\n"); } int main() { int T; scanf("%d", &T); while (T--) { input(); solve(); } return 0; }
//邻接表 #include <cstdio> #include <cstring> using namespace std; int p, n; const int maxn = 310; struct edge { int v, next; }e[30010]; int linker[maxn]; bool vis[maxn]; int k, head[30010]; void adde(int u, int v) { e[k].v = v; e[k].next = head[u]; head[u] = k++; } void input() { scanf("%d%d", &p, &n); int t, v; k = 1; memset(head, -1, sizeof(head)); for (int i = 1; i <= p; i++) { scanf("%d", &t); while (t--) { scanf("%d", &v); adde(i, v); } } } bool dfs(int u) { for (int i = head[u]; i != -1; i = e[i].next) { int v = e[i].v; if (!vis[v]) { vis[v] = 1; if (linker[v] == -1 || dfs(linker[v])) { linker[v] = u; return 1; } } } return 0; } int hungary() { int res = 0; memset(linker, -1, sizeof(linker)); for (int u = 1; u <= p; u++) { memset(vis, 0, sizeof(vis)); if (dfs(u)) res++; } return res; } void solve() { if (hungary() == p) printf("YES\n"); else printf("NO\n"); } int main() { int T; scanf("%d", &T); while (T--) { input(); solve(); } return 0; }