• 无源汇上下界可行流(多校7)


    http://acm.hdu.edu.cn/showproblem.php?pid=4940

    Destroy Transportation system

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
    Total Submission(s): 298    Accepted Submission(s): 184


    Problem Description
    Tom is a commander, his task is destroying his enemy’s transportation system.

    Let’s represent his enemy’s transportation system as a simple directed graph G with n nodes and m edges. Each node is a city and each directed edge is a directed road. Each edge from node u to node v is associated with two values D and B, D is the cost to destroy/remove such edge, B is the cost to build an undirected edge between u and v.

    His enemy can deliver supplies from city u to city v if and only if there is a directed path from u to v. At first they can deliver supplies from any city to any other cities. So the graph is a strongly-connected graph.

    He will choose a non-empty proper subset of cities, let’s denote this set as S. Let’s denote the complement set of S as T. He will command his soldiers to destroy all the edges (u, v) that u belongs to set S and v belongs to set T.

    To destroy an edge, he must pay the related cost D. The total cost he will pay is X. You can use this formula to calculate X:

    After that, all the edges from S to T are destroyed. In order to deliver huge number of supplies from S to T, his enemy will change all the remained directed edges (u, v) that u belongs to set T and v belongs to set S into undirected edges. (Surely, those edges exist because the original graph is strongly-connected)

    To change an edge, they must remove the original directed edge at first, whose cost is D, then they have to build a new undirected edge, whose cost is B. The total cost they will pay is Y. You can use this formula to calculate Y:

    At last, if Y>=X, Tom will achieve his goal. But Tom is so lazy that he is unwilling to take a cup of time to choose a set S to make Y>=X, he hope to choose set S randomly! So he asks you if there is a set S, such that Y<X. If such set exists, he will feel unhappy, because he must choose set S carefully, otherwise he will become very happy.
     

    Input
    There are multiply test cases.

    The first line contains an integer T(T<=200), indicates the number of cases.

    For each test case, the first line has two numbers n and m.

    Next m lines describe each edge. Each line has four numbers u, v, D, B.
    (2=<n<=200, 2=<m<=5000, 1=<u, v<=n, 0=<D, B<=100000)

    The meaning of all characters are described above. It is guaranteed that the input graph is strongly-connected.
     

    Output
    For each case, output "Case #X: " first, X is the case number starting from 1.If such set doesn’t exist, print “happy”, else print “unhappy”.
     

    Sample Input
    2 3 3 1 2 2 2 2 3 2 2 3 1 2 2 3 3 1 2 10 2 2 3 2 2 3 1 2 2
     

    Sample Output
    Case #1: happy Case #2: unhappy
    Hint
    In first sample, for any set S, X=2, Y=4. In second sample. S= {1}, T= {2, 3}, X=10, Y=4.

    题意:给出一个有向强连通图,每条边有两个值分别是破坏该边的代价和把该边建成无向边的代价(建立无向边的前提是删除该边)问是否存在一个集合S,和一个集合的补集T,破坏所有S集合到T集合的边代价和是X,然后修复T到S的边为无向边代价和是Y,满足Y<X;满足输出unhappy,否则输出happy;

    分析:首先可以把每条边的权值做一下变换,即破坏有向边的权值A=d,和建立无向边的权值B=b+d;

    官方题解:




    程序:

    #include"string.h"
    #include"stdio.h"
    #include"iostream"
    #include"queue"
    #define inf 100000000
    #include"math.h"
    #define M 333
    #define eps 1e-5
    using namespace std;
    struct node
    {
        int u,v,w,c,next;
    }edge[40009];
    int t,head[M],dis[M],work[M];
    void init()
    {
        t=0;
        memset(head,-1,sizeof(head));
    }
    void add(int u,int v,int w,int c)
    {
        edge[t].u=u;
        edge[t].v=v;
        edge[t].w=w;
        edge[t].c=c;
        edge[t].next=head[u];
        head[u]=t++;
    
        edge[t].u=v;
        edge[t].v=u;
        edge[t].w=0;
        edge[t].c=c;
        edge[t].next=head[v];
        head[v]=t++;
    }
    int bfs(int start,int endl)
    {
        queue<int>q;
        memset(dis,-1,sizeof(dis));
        dis[start]=0;
        q.push(start);
        while(!q.empty())
        {
            int u=q.front();
            q.pop();
            for(int i=head[u];i!=-1;i=edge[i].next)
            {
                int v=edge[i].v;
                if(edge[i].w&&dis[v]==-1)
                {
                    dis[v]=dis[u]+1;
                    q.push(v);
                    if(v==endl)
                        return 1;
                }
            }
        }
        return 0;
    }
    int dfs(int u,int a,int T)
    {
        if(u==T)
            return a;
        for(int &i=work[u];i!=-1;i=edge[i].next)
        {
            int v=edge[i].v;
            if(edge[i].w&&dis[v]==dis[u]+1)
            {
                int tt=dfs(v,min(edge[i].w,a),T);
                if(tt)
                {
                    edge[i].w-=tt;
                    edge[i^1].w+=tt;
                    return tt;
                }
            }
        }
        return 0;
    }
    int Dinic(int S,int T)
    {
        int ans=0;
        while(bfs(S,T))
        {
            memcpy(work,head,sizeof(head));
            while(int tt=dfs(S,inf,T))
                ans+=tt;
        }
        return ans;
    }
    int main()
    {
        int T,kk=1;
        cin>>T;
        while(T--)
        {
            int n,m;
            scanf("%d%d",&n,&m);
            init();
            int st=0;
            int sd=n+1;
            int sum=0;
            while(m--)
            {
                int a,b,c,d;
                scanf("%d%d%d%d",&a,&b,&c,&d);
                sum+=c;
                add(a,b,d,c+d);
                add(a,sd,c,c);
                add(st,b,c,c);
            }
            int ans=Dinic(st,sd);
            printf("Case #%d: ",kk++);
            if(sum!=ans)
            {
                printf("unhappy
    ");
            }
            else
                printf("happy
    ");
        }
        return 0;
    }
    


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  • 原文地址:https://www.cnblogs.com/mypsq/p/4348180.html
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