HDU 4940 Destroy Transportation system(无源汇上下界网络流)

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

 

Sample Output

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

给你N个点M条边强连通的有向简单图，D代表删掉这个边的花费，D+B代表重建为双向边的花费，让你选择一个集合S，其余的点在T集合，X为u在S集合v在T集合的所有边的D之和，Y为u在T集合v在S集合的所有边的D+B之和，求是否存在一个集合S，使得X>Y，若存在输出unhappy，否则输出happy

题解

无源汇上下界网络流，下界D，上界D+B，判断是否存在可行流

若存在，则说明对于任意集合S，流出的流量=流入的流量，X<=流出的流量<=Y

建图每条边建为自由流（u，v，B）

对于每个点，设M为总流入-总流出

若M>0，则建（S，i，M）说明i需要多流出M

若M<0，则建（i，T，M）说明i需要多流入M

最后判断与S连的边是否全满流

代码

#include<bits/stdc++.h>
using namespace std;

const int maxn=1e5+5;
const int maxm=2e5+5;
const int INF=0x3f3f3f3f;

int TO[maxm],CAP[maxm],NEXT[maxm],tote;
int FIR[maxn],gap[maxn],cur[maxn],d[maxn],q[400000];
int n,m,S,T;

void add(int u,int v,int cap)
{
    TO[tote]=v;
    CAP[tote]=cap;
    NEXT[tote]=FIR[u];
    FIR[u]=tote++;
    
    TO[tote]=u;
    CAP[tote]=0;
    NEXT[tote]=FIR[v];
    FIR[v]=tote++;
}
void bfs()
{
    memset(gap,0,sizeof gap);
    memset(d,0,sizeof d);
    ++gap[d[T]=1];
    for(int i=1;i<=n;++i)cur[i]=FIR[i];
    int head=1,tail=1;
    q[1]=T;
    while(head<=tail)
    {
        int u=q[head++];
        for(int v=FIR[u];v!=-1;v=NEXT[v])
            if(!d[TO[v]])
                ++gap[d[TO[v]]=d[u]+1],q[++tail]=TO[v];
    }
}
int dfs(int u,int fl)
{
    if(u==T)return fl;
    int flow=0;
    for(int &v=cur[u];v!=-1;v=NEXT[v])
        if(CAP[v]&&d[u]==d[TO[v]]+1)
        {
            int Min=dfs(TO[v],min(fl,CAP[v]));
            flow+=Min,fl-=Min,CAP[v]-=Min,CAP[v^1]+=Min;
            if(!fl)return flow;
        }
    if(!(--gap[d[u]]))d[S]=n+1;
    ++gap[++d[u]],cur[u]=FIR[u];
    return flow;
}
int ISAP()
{
    bfs();
    int ret=0;
    while(d[S]<=n)ret+=dfs(S,INF);
    return ret;
}
void init()
{
    tote=0;
    memset(FIR,-1,sizeof FIR);
}
int in[maxn];
int main()
{
    int t;
    scanf("%d",&t);
    for(int ca=1;ca<=t;ca++)
    {
        init();
        memset(in,0,sizeof in);
        scanf("%d%d",&n,&m);
        for(int i=0,u,v,d,b;i<m;i++)
        {
            scanf("%d%d%d%d",&u,&v,&d,&b);
            add(u,v,b);
            in[u]-=d;
            in[v]+=d;
        }
        S=n+1,T=S+1,n+=2;
        int sum=0;
        for(int i=1;i<=n;i++)
            if(in[i]>0)
            {
                add(S,i,in[i]);
                sum+=in[i];
            }
            else if(in[i]<0)
                add(i,T,-in[i]);
        printf("Case #%d: %s
",ca,ISAP()==sum?"happy":"unhappy");
    }
    return 0;
}