二分图最小点权覆盖（输出覆盖的点）poj2025

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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

Sample Output

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;
}