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\n");
        }
        else
            printf("happy\n");
    }
    return 0;
}