hdoj 1845 Jimmy’s Assignment 【HK】

最新推荐文章于 2020-07-28 17:32:20 发布

笑着走完自己的路

最新推荐文章于 2020-07-28 17:32:20 发布

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分类专栏：最大匹配(权值匹配)、独立集、团 && 最小路径、点覆盖

本文链接：https://blog.youkuaiyun.com/chenzhenyu123456/article/details/49201833

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本文探讨了一个关于最大匹配的问题，通过实例介绍了如何解决这种特定类型的二分图匹配问题。详细解释了最大匹配的概念、输入输出规范，并提供了一种高效算法——HK算法来求解。通过样例输入和输出，清晰地展示了算法的应用过程和结果。

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Jimmy’s Assignment

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 1304 Accepted Submission(s): 527

Problem 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

Sample Output

2
2

以前没A的题，今天KO了。

题意：给你n个点和n*3/2条边，求出最大匹配。

HK，注意设置numx = numy = n后，最大匹配结果除2。

AC代码：

#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
#define MAXN 5000+10
using namespace std;
int mx[MAXN], my[MAXN];//记录X和Y集元素的匹配点
int dx[MAXN], dy[MAXN];//记录距离
bool used[MAXN];//标记是否使用
vector<int> G[MAXN];//存储二分图
int n, m;//点数 边数
int numx, numy;//记录X集元素个数 Y集元素个数
int DFS(int u)
{
    for(int i = 0; i < G[u].size(); i++)
    {
        int v = G[u][i];
        if(!used[v] && dy[v] == dx[u] + 1)
        {
            used[v] = true;
            if(my[v] == -1 || DFS(my[v]))
            {
                my[v] = u;
                mx[u] = v;
                return 1;
            }
        }
    }
    return 0;
}
void HK_match()
{
    memset(mx, -1, sizeof(mx));
    memset(my, -1, sizeof(my));
    int ans = 0;
    while(1)
    {
        bool flag = false;//标记是否找到可增广路径
        memset(dx, 0, sizeof(dx));
        memset(dy, 0, sizeof(dy));
        queue<int> Q;
        for(int i = 1; i <= numx; i++)
            if(mx[i] == -1) Q.push(i);
        while(!Q.empty())//寻找增广路
        {
            int u = Q.front(); Q.pop();
            for(int i = 0; i < G[u].size(); i++)
            {
                int v = G[u][i];
                if(!dy[v])
                {
                    dy[v] = dx[u] + 1;
                    if(my[v] == -1)//未匹配 说明找到增广路
                        flag = true;
                    else
                    {
                        dx[my[v]] = dx[u] + 1;
                        Q.push(my[v]);
                    }
                }
            }
        }
        if(!flag)//未找到增广路
            break;
        memset(used, false, sizeof(used));
        for(int i = 1; i <= numx; i++)
            if(mx[i] == -1)
                ans += DFS(i);
    }
    printf("%d\n", ans / 2);
}
void getMap()
{
    scanf("%d", &n);
    for(int i = 1; i <= n; i++)
        G[i].clear();
    m = 3*n/2;
    numx = numy = n;
    while(m--)
    {
        int a, b;
        scanf("%d%d", &a, &b);
        G[a].push_back(b);
        G[b].push_back(a);
    }
}
int main()
{
    int t;
    scanf("%d", &t);
    while(t--)
    {
        getMap();
        HK_match();
    }
    return 0;
}