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HDU - 1845 Jimmy’s Assignment （二分匹配）

2024-07-12 06:53:02   220人阅读

Description

Jimmy is studying Advanced Graph Algorithms at his university. His most recent assignment is to find a maximum matching in a special kind of graph. This graph is undirected, has N vertices and each vertex has degree 3. Furthermore, the graph is 2-edge-connected (that is, at least 2 edges need to be removed in order to make the graph disconnected). A matching is a subset of the graph’s edges, such that no two edges in the subset have a common vertex. A maximum matching is a matching having the maximum cardinality.
  Given a series of instances of the special graph mentioned above, find the cardinality of a maximum matching for each instance.
 

Input

The first line of input contains an integer number T, representing the number of graph descriptions to follow. Each description contains on the first line an even integer number N (4<=N<=5000), representing the number of vertices. Each of the next 3*N/2 lines contains two integers A and B, separated by one blank, denoting that there is an edge between vertex A and vertex B. The vertices are numbered from 1 to N. No edge may appear twice in the input.
 

Output

For each of the T graphs, in the order given in the input, print one line containing the cardinality of a maximum matching.
 

Sample Input

2
4
1 2
1 3
1 4
2 3
2 4
3 4
4
1 2
1 3
1 4
2 3
2 4
3 4
 

Sample Output

2
2
 

Source

Politehnica University of Bucharest Local Team Contest 2007

题意：给你双向边，求最多留下多少条边使得每条边都没有共有顶点

思路：二分图匹配的定义，对于双向的要/2，用vector会超时，要用邻接表

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <iostream>
using namespace std;
const int MAXN = 5010;
const int MAXM = 50010;

struct Edge {
	int to, next;
} edge[MAXM];
int head[MAXN], tot;
int linker[MAXN];
bool used[MAXN];
int n, m;

void init() {
	tot = 0;
	memset(head,-1,sizeof(head));
}

void addEdge(int u, int v) {
	edge[tot].to = v; edge[tot].next = head[u];
	head[u] = tot++;
}

bool dfs(int u) {
	for (int i = head[u]; i != -1; i = edge[i].next) {
		int v = edge[i].to;
		if (!used[v]) {
			used[v] = true;
			if (linker[v] == -1 || dfs(linker[v])) {
				linker[v] = u;
				return true;
			}
		}
	}
	return false;
}

int solve() {
	int ans = 0;
	memset(linker, -1, sizeof(linker));
	for (int i = 0; i < n; i++) {
		memset(used, false, sizeof(used));
		if (dfs(i))
			ans++;
	}
	return ans;
}

int main() {
	int t;
	scanf("%d",&t);
	while (t--) {
		scanf("%d", &n);
		m = n*3/2;
		int u,v;
		init();
		while (m--) {
			scanf("%d%d", &u, &v);
			u--; v--;
			addEdge(u,v);
			addEdge(v,u);
		}
		printf("%d\n", solve()/2);
	}
	return 0;
}


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