二分图最小点权覆盖(输出覆盖的点)poj2025

本文深入探讨了游戏开发领域的关键技术,包括游戏引擎、编程语言、硬件优化等,并重点阐述了AI音视频处理的应用场景和实现方法,如语义识别、物体检测、语音变声等。通过实例分析,揭示了现代游戏与AI融合的趋势。

Language:
Destroying The Graph
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 7319 Accepted: 2321 Special Judge

Description

Alice and Bob play the following game. First, Alice draws some directed graph with N vertices and M arcs. After that Bob tries to destroy it. In a move he may take any vertex of the graph and remove either all arcs incoming into this vertex, or all arcs outgoing from this vertex.  
Alice assigns two costs to each vertex: Wi +  and Wi -. If Bob removes all arcs incoming into the i-th vertex he pays Wi +  dollars to Alice, and if he removes outgoing arcs he pays Wi -  dollars.  
Find out what minimal sum Bob needs to remove all arcs from the graph.

Input

Input file describes the graph Alice has drawn. The first line of the input file contains N and M (1 <= N <= 100, 1 <= M <= 5000). The second line contains N integer numbers specifying Wi +. The third line defines Wi -  in a similar way. All costs are positive and do not exceed 10 6  . Each of the following M lines contains two integers describing the corresponding arc of the graph. Graph may contain loops and parallel arcs.

Output

On the first line of the output file print W --- the minimal sum Bob must have to remove all arcs from the graph. On the second line print K --- the number of moves Bob needs to do it. After that print K lines that describe Bob's moves. Each line must first contain the number of the vertex and then '+' or '-' character, separated by one space. Character '+' means that Bob removes all arcs incoming into the specified vertex and '-' that Bob removes all arcs outgoing from the specified vertex.

Sample Input

3 6
1 2 3
4 2 1
1 2
1 1
3 2
1 2
3 1
2 3

Sample Output

5
3
1 +
2 -
2 +

转化成最小点权覆盖见图的方法,参见论文《最小割模型在信息学竞赛中的应用》

这里想说一下如何输出这些点,前几天在uva上遇到过这种题,没仔细看,现在总结一下

首先从X的未匹配点出发dfs直到不能再走,这样X集合未标记的点和Y集合标记了的点就是可以覆盖所有边的点

uva11419

#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<vector>
#include<cmath>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<algorithm>
using namespace std;
const int maxn=210;
const int INF=1000000000;
int A[maxn],B[maxn];
int N,M,s,t,nn;
int vis[maxn],pre[maxn],cnt;
int dis[maxn],gap[maxn],cur[maxn],head[maxn],biao[maxn];
//vector<int> ans;

struct node
{
    int u,v,f,next;
}edge[15000];
void add_edge(int x,int y,int f)
{
    edge[cnt].u=x;
    edge[cnt].v=y;
    edge[cnt].f=f;
    edge[cnt].next=head[x];
    head[x]=cnt++;
    edge[cnt].u=y;
    edge[cnt].v=x;
    edge[cnt].f=0;
    edge[cnt].next=head[y];
    head[y]=cnt++;
}
int SAP(int st,int en)
{
    for(int i=0;i<=nn;i++)
    {
        cur[i]=head[i];
        dis[i]=gap[i]=0;
    }
    int u=0;
    int flow=0,aug=INF;
    gap[st]=nn;
    u=pre[st]=st;
    bool flag;
    while(dis[st]<nn)
    {
        flag=0;
        for(int &j=cur[u];j!=-1;j=edge[j].next)
        {
            int v=edge[j].v;
            if(edge[j].f>0&&dis[u]==dis[v]+1)
            {
                flag=1;
                if(edge[j].f<aug)aug=edge[j].f;
                pre[v]=u;
                u=v;
                if(u==en)
                {
                    flow+=aug;
                    while(u!=st)
                    {
                        u=pre[u];
                        edge[cur[u]].f-=aug;
                        edge[cur[u]^1].f+=aug;
                    }
                    aug=INF;
                }
                break;
            }
        }

        if(flag)continue;
        int mindis=nn;
        for(int j=head[u];j!=-1;j=edge[j].next)
        {
            int v=edge[j].v;
            if(dis[v]<mindis&&edge[j].f>0)
            {
                mindis=dis[v];
                cur[u]=j;
            }
        }
        if((--gap[dis[u]])==0)break;
        gap[dis[u]=mindis+1]++;
        u=pre[u];
    }
    return flow;
}
void DFS(int u)
{
    for(int i=head[u];i!=-1;i=edge[i].next)
    {
        int v=edge[i].v;
        if(biao[v])continue;
        if(edge[i].f>0)
        {
            biao[v]=1;
            DFS(v);
        }
    }
}
int main()
{
    while(scanf("%d%d",&N,&M)!=EOF)
    {
        for(int i=1;i<=N;i++)scanf("%d",&A[i]);
        for(int i=1;i<=N;i++)scanf("%d",&B[i]);
        s=0,t=N*2+1,nn=t+1;
        cnt=0;
        memset(head,-1,sizeof(head));
        for(int i=1;i<=N;i++)
        {
            add_edge(s,i,B[i]);
            add_edge(i+N,t,A[i]);
        }
        while(M--)
        {
            int u,v;
            scanf("%d%d",&u,&v);
            add_edge(u,v+N,INF);
        }
        memset(biao,0,sizeof(biao));
        int ans=SAP(s,t);
        DFS(s);
        printf("%d\n",ans);
        ans=0;
        for(int i=1;i<=2*N;i++)
        {
            if(!biao[i]&&i<=N)ans++;
            else if(biao[i]&&i>N)ans++;
        }
        printf("%d\n",ans);
        for(int i=1;i<=N;i++)
        {
            if(!biao[i])printf("%d -\n",i);
            if(biao[i+N])printf("%d +\n",i);
        }
    }
    return 0;
}




### 关于二分图小顶覆盖的算法实现 #### 1. 小顶覆盖的概念 小顶覆盖是指在一个二分图中选取尽可能少的节,使得这些节能够覆盖所有的边。换句话说,对于每一条边 (u, v),至少有一个端 u 或 v 被选入覆盖集中。 根据 König 定理,在任意无向二分图中,小顶覆盖的数量等于该图的大匹配数量[^1]。 --- #### 2. 算法原理 为了求解二分图小顶覆盖,通常采用 **匈牙利算法** 来计算大匹配数。具体过程如下: - 构建一个二分图 G(X,Y,E),其中 X 和 Y 是两个互不相交的节集合,E 表示连接它们的边。 - 使用匈牙利算法找到二分图大匹配 M。 - 基于大匹配的结果,通过以下方式构造小顶覆盖: - 将左侧未被匹配的节加入到覆盖集中; - 对右侧已被匹配的节也加入到覆盖集中。 终得到的覆盖集大小即为小顶覆盖的数量[^2]。 --- #### 3. 实现代码 以下是基于 Python 的实现代码,利用匈牙利算法完成二分图小顶覆盖的计算: ```python from collections import defaultdict def hungarian_algorithm(graph, n, m): """ 匈牙利算法用于寻找二分图大匹配 :param graph: 邻接表表示的二分图 {X -> [Y]} :param n: 左侧节数目 :param m: 右侧节数目 :return: 大匹配结果 """ match_y = [-1] * m # 记录右侧节对应的匹配关系 visited = None # 当前轮次访问标记 def dfs(u): for v in graph[u]: if not visited[v]: visited[v] = True if match_y[v] == -1 or dfs(match_y[v]): match_y[v] = u return True return False matching_count = 0 for i in range(n): visited = [False] * m if dfs(i): matching_count += 1 return matching_count, match_y def min_vertex_cover(graph, n, m): """ 求解二分图小顶覆盖 :param graph: 邻接表表示的二分图 {X -> [Y]} :param n: 左侧节数目 :param m: 右侧节数目 :return: 小顶覆盖集合 """ max_matching, match_y = hungarian_algorithm(graph, n, m) cover_x = set() # 左侧需要覆盖的节 cover_y = set() # 右侧需要覆盖的节 unmatched_in_x = set(range(n)) # 初始认为所有左侧节都未匹配 matched_in_y = set() for y in range(m): # 找出右侧已匹配的节 if match_y[y] != -1: unmatched_in_x.discard(match_y[y]) # 如果某个左侧节参与了匹配,则移除 matched_in_y.add(y) cover_x.update(unmatched_in_x) # 添加左侧未匹配的节 cover_y.update(set(range(m)) - matched_in_y) # 添加右侧未匹配的节 return list(cover_x), list(cover_y) # 测试用例 if __name__ == "__main__": # 输入邻接表形式的二分图 graph = { 0: [0, 1], 1: [0, 2], 2: [1, 3] } n = 3 # 左侧节数 m = 4 # 右侧节数 result_x, result_y = min_vertex_cover(graph, n, m) print(f"左侧需覆盖的节: {result_x}") print(f"右侧需覆盖的节: {result_y}") ``` 上述代码实现了二分图小顶覆盖功能,核心部分依赖匈牙利算法来获取大匹配,并据此推导出覆盖所需的节集合[^3]。 --- #### 4. 应用实例 考虑 POJ 3041 Asteroids 这道题目,其本质是一个二分图小顶覆盖问题。给定一组障碍物坐标,将其转化为二分图模型后,可以通过以上方法高效解决。输出的是满足条件的小射击次数[^4]。 ---
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值