[toj1047]【二分图最大独立集】Girls and Boys

最新推荐文章于 2019-05-09 22:40:37 发布

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#匈牙利算法 #二分图匹配 #二分图最大独立集

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本文介绍了一种通过求解最大匹配来找到图中最大独立集的方法，并提供了一个具体的实现案例。针对大学校园内学生间的浪漫关系研究问题，利用二分图特性解决了最大独立集的求解难题。

1047. Girls and Boys

Time Limit: 5.0 Seconds Memory Limit: 65536K
Total Runs: 1173 Accepted Runs: 385

the second year of the university somebody started a study on the romantic relations between the students. The relation "romantically involved" is defined between one girl and one boy. For the study reasons it is necessary to find out the maximum set satisfying the condition: there are no two students in the set who have been "romantically involved". The result of the program is the number of students in such a set.

The input contains several data sets in text format. Each data set represents one set of subjects of the study, with the following description:

the number of students
the description of each student, in the following format
student_identifier:(number_of_romantic_relations) student_identifier1 student_identifier2 student_identifier3 ...
or
student_identifier:(0)

The student_identifier is an integer number between 0 and n-1, for n subjects.
For each given data set, the program should write to standard output a line containing the result.

Sample Input

7
0: (3) 4 5 6
1: (2) 4 6
2: (0)
3: (0)
4: (2) 0 1
5: (1) 0
6: (2) 0 1
3
0: (2) 1 2
1: (1) 0
2: (1) 0

Sample Output

5
2

Source: Southeastern European 2000

Show Code - Run ID 1488836
Submit Time: 2014-12-15 18:35:16     Language: GNU C++     Result: Accepted
    Pid: 1047     Time: 0.11 sec.     Memory: 956 K.     Code Length: 1.2 K.

/*
二分图：最大独立集=|V|-最小顶点覆盖=|V|-最大匹配
*/ 
#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
using namespace std;

const int MAX = 512;
int match[MAX];
bool vis[MAX];
vector<int> G[MAX];
int n;

bool dfs(int u) {
	for (vector<int>::iterator it =  G[u].begin(); it != G[u].end(); ++it) {
		if (!vis[*it]) {
			vis[*it] = true;
			if (match[*it] == -1 || dfs(match[*it])) {
				match[*it] = u;
				return true;
			}
		}
	}
	return false;
}

inline int read() {
	char ch;
	while ((ch = getchar()) < '0' || ch > '9');
	int x = ch - '0';
	while ((ch = getchar()) >= '0' && ch <= '9') {
		x = (x << 3) + (x << 1) + ch - '0';
	}
	return x;
}

int main() {
	while (~scanf(" %d", &n)) {
		for (int i = 0; i < n; ++i) G[i].clear();
		
		int id, m, x;
		for (int i = 0; i < n; ++i) {
			id = read();
			m = read();
			while (m--) {
				x = read();
				G[id].push_back(x);
			}
		}
		
		memset(match, -1, sizeof(match));
		int tot = 0;
		for (int i = 0; i < n; ++i) {
			memset(vis, false, sizeof(vis));
			if (dfs(i)) ++tot;
		}
		printf("%d\n", n - (tot >> 1));
	}
	return 0;
}