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.

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.

同上一道题，不过在我不知道这题要用最大流来做的情况下我是不会想到的：

关键是要构造出不等式，而且把不等式对应到可行流。

#include<cstdio>

#include<cstdlib>

#include<iostream>

#include<cstring>

#include<algorithm>

using namespace std;

const int maxn=4;

const int inf=;

int Laxt[maxn],Next[maxn],To[maxn],Cap[maxn],cnt;

int dis[maxn],nd[maxn],S,T,num,ans,q[maxn],qnum[maxn],top;

void init()

{

    cnt=;ans=num=top=;

    memset(Laxt,,sizeof(Laxt));

    memset(dis,,sizeof(dis));

    memset(nd,,sizeof(nd));

}

int add(int u,int v,int c)

{

    Next[++cnt]=Laxt[u];

    Laxt[u]=cnt;

    To[cnt]=v;

    Cap[cnt]=c;

    Next[++cnt]=Laxt[v];

    Laxt[v]=cnt;

    To[cnt]=u;

    Cap[cnt]=;

}

int sap(int u,int flow)

{

    if(u==T||flow==) return flow;

    int delta=,tmp;

    for(int i=Laxt[u];i;i=Next[i]){

        int v=To[i];

        if(dis[v]+==dis[u]&&Cap[i]>){

            tmp=sap(v,min(Cap[i],flow-delta));

            delta+=tmp;

            Cap[i]-=tmp;

            Cap[i^]+=tmp;

            if(flow==delta||dis[]>=T) return delta;

        }

    }

    nd[dis[u]]--;

    if(nd[dis[u]]==) dis[]=T;

    nd[++dis[u]]++;

    return delta;

}

int main()

{

    int Case,n,i,j,m,u,v,x,y,k=;

    scanf("%d",&Case);

    while(Case--){

        init();

        scanf("%d%d",&n,&m);

        S=;T=n+;

        for(i=;i<=m;i++){

            scanf("%d%d%d%d",&u,&v,&x,&y);

            u++;v++;num+=x;

            add(u,v,y);

            q[++top]=cnt;

            qnum[top]=x;

            add(S,v,x);

            add(u,T,x);

        }

        while(dis[S]<T) {

           ans+=sap(S,inf);

        }

        printf("Case #%d: ",++k);

        if(num!=ans) printf("unhappy\n");

        else printf("happy\n");

    }

    return ;

}

（希望多遇到几个这样的模型，然后好好理解一下）