hdu1083【二分图】【最大匹配】
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Courses
Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7057 Accepted Submission(s): 3458
Problem 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
Your program should read sets of data from a text file. The first line of the input file 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’ll 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.
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.
An example of program input and 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
一共有N个学生跟P门课程,一个学生可以任意选一门或多门课,问是否达成:
1.每个学生代表的都是不同的课(如果一个学生选修的那门课,那么他就可以代表这门课)
2.每门课都有一个代表
什么是最大匹配问题和二分图?
请点击二分图介绍
代码:
#include <iostream>#include <string>#include <cstring>#include <cstdio>#include <cmath>#include <cstdlib>#include <algorithm>#include <queue>#include <map>#include <sstream>#define Maxn 100005using namespace std;bool edge[500][500];int x, y;int cx[305], cy[305];bool vis[305];int path(int u){ int v; for (v = x + 1; v <= x + y; v++) { if (edge[u][v] && !vis[v]) { vis[v] = 1; if (cy[v] == -1 || path(cy[v])) { cx[u] = v; cy[v] = u; return 1; } } } return 0;}int main(){ int T, n, a; cin >> T; while (T--) { memset(cx, -1, sizeof(cx)); memset(cy, -1, sizeof(cy)); memset(edge, 0, sizeof(edge)); cin >> x >> y; for (int i = 1; i <= x; i++) { scanf("%d", &n); while (n--) { scanf("%d", &a); edge[a + x][i] = 1; edge[i][a + x] = 1; } } int ans = 0; for (int i = 1; i <= x; i++) { memset(vis, 0, sizeof(vis)); ans += path(i); } if (ans == x) printf("YES\n"); else printf("NO\n"); }}
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