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
Hint
In first sample, for any set S, X=2, Y=4. In second sample. S= {1}, T= {2, 3}, X=10, Y=4. Author
UESTC
Source
感觉官方题解说的好明白==
说一下对于有上下限制的网络流的更深一些点的理解：
网络流总结篇 的第16题的图是这个意思，但是建边的原理没说明白
给定的是A->B的范围是[c,d]，对于点B而言，他的流量必须有C，人为补充的是d-c的可行流，对于点A来说，流出的流量至少是C，d-c的部分是可行流。
#include <iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int mm=1000000;
const int mn=505*505*3;
const int oo=1000000000;
int node,src,dest,edge;
int reach[mm],flow[mm],nxt[mm];
int head[mn],work[mn],dis[mn],q[mn];
inline int min(int a,int b)
{
return a<b?a:b;
}
inline void prepare(int _node,int _src,int _dest)
{
node=_node,src=_src,dest=_dest;
for(int i=0;i<node;++i)head[i]=-1;
edge=0;
}
inline void addedge(int u,int v,int c1,int c2=0)
{
reach[edge]=v,flow[edge]=c1,nxt[edge]=head[u],head[u]=edge++;
reach[edge]=u,flow[edge]=c2,nxt[edge]=head[v],head[v]=edge++;
}
bool Dinic_bfs()
{
int i,u,v,l,r=0;
for(i=0;i<node;++i)dis[i]=-1;
dis[q[r++]=src]=0;
for(l=0;l<r;++l)
for(i=head[u=q[l]];i>=0;i=nxt[i])
if(flow[i]&&dis[v=reach[i]]<0)
{
dis[q[r++]=v]=dis[u]+1;
if(v==dest)return 1;
}
return 0;
}
int Dinic_dfs(int u,int exp)
{
if(u==dest)return exp;
for(int &i=work[u],v,tmp;i>=0;i=nxt[i])
if(flow[i]&&dis[v=reach[i]]==dis[u]+1&&(tmp=Dinic_dfs(v,min(exp,flow[i])))>0)
{
flow[i]-=tmp;
flow[i^1]+=tmp;
return tmp;
}dis[u]--;
return 0;
}
int Dinic_flow()
{
int i,ret=0,delta;
while(Dinic_bfs())
{
for(i=0;i<node;++i)work[i]=head[i];
while(delta=Dinic_dfs(src,oo))ret+=delta;
}
return ret;
}
int Edge[40000],Flow[40000];
int main()
{
// freopen("cin.txt","r",stdin);
int n,m,t,cas=1;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&m);
prepare(n+2,0,n+1);
int sum=0;
for(int i=0;i<m;i++)
{
int a,b,c,d;
scanf("%d%d%d%d",&a,&b,&c,&d);
d+=c;
sum+=c;
addedge(src,b,c);
addedge(a,dest,c);
Edge[i]=edge;
Flow[i]=c;
addedge(a,b,d-c);
}
printf("Case #%d: ",cas++);
if(sum!=Dinic_flow())
{
puts("unhappy");
}
else
{
puts("happy");
}
}
return 0;
}
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