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.
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.
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
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=40000; const int inf=1000000000; 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=1;ans=num=top=0; memset(Laxt,0,sizeof(Laxt)); memset(dis,0,sizeof(dis)); memset(nd,0,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]=0; } int sap(int u,int flow) { if(u==T||flow==0) return flow; int delta=0,tmp; for(int i=Laxt[u];i;i=Next[i]){ int v=To[i]; if(dis[v]+1==dis[u]&&Cap[i]>0){ tmp=sap(v,min(Cap[i],flow-delta)); delta+=tmp; Cap[i]-=tmp; Cap[i^1]+=tmp; if(flow==delta||dis[1]>=T) return delta; } } nd[dis[u]]--; if(nd[dis[u]]==0) dis[1]=T; nd[++dis[u]]++; return delta; } int main() { int Case,n,i,j,m,u,v,x,y,k=0; scanf("%d",&Case); while(Case--){ init(); scanf("%d%d",&n,&m); S=1;T=n+2; for(i=1;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 "); else printf("happy "); } return 0; }
（希望多遇到几个这样的模型，然后好好理解一下）