【HDU 4940】Destroy Transportation system（无源无汇带上下界可行流）

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.

Sample Input

Sample Output

Source

每个点单独作为S集合时，如果存在满足Y<X，就输出unhappy。否则输出happy。

#include <cstdio>
#include <cstring>
#define N 205
#define sf(x) scanf("%d",&x)
int x[N],y[N];
int main(){
int t;
sf(t);
for(int cas=;cas<=t;cas++){
memset(x,,sizeof x);
memset(y,,sizeof y);
printf("Case #%d: ",cas);
int n,m;
sf(n);
sf(m);
for(int i=;i<=m;i++){
int u,v,d,b;
sf(u);sf(v);sf(d);sf(b);
x[u]+=d;
y[v]+=d+b;
}
int ok=;
for(int i=;i<=n;i++)
if(x[i]>y[i])ok=;
if(ok)puts("happy");
else puts("unhappy");
}
}

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
#define N 500
#define M 200001
#define inf 0x3f3f3f3f
struct edge{
int to,next,cap,flow;
}e[M];
int head[N],cnt;
int gap[N],dep[N],cur[N];
void init(){
cnt=;
memset(head, -, sizeof head);
}
void add(int u,int v,int w,int rw=){
e[cnt]=(edge){v,head[u],w,};
head[u]=cnt++;
e[cnt]=(edge){u,head[v],rw,};
head[v]=cnt++;
}
int q[N];
void bfs(int st,int ed){
memset(dep,-,sizeof dep);
memset(gap,,sizeof gap);
gap[]=;
int front=,rear=;
dep[ed]=;
q[rear++]=ed;
while(front!=rear){
int u=q[front++];
for(int i=head[u];~i;i=e[i].next){
int v=e[i].to;
if(dep[v]!=-)continue;
q[rear++]=v;
dep[v]=dep[u]+;
gap[dep[v]]++;
}
}
}
int s[N];
int sap(int st,int ed,int n){
bfs(st,ed);
memcpy(cur,head,sizeof head);
int top=;
int u=st;
int ans=;
while(dep[st]<n){
if(u==ed){
int Min=inf;
int inser;
for(int i=;i<top;i++)
if(Min>e[s[i]].cap-e[s[i]].flow){
Min=e[s[i]].cap-e[s[i]].flow;
inser=i;
}
for(int i=;i<top;i++){
e[s[i]].flow+=Min;
e[s[i]^].flow-=Min;
}
ans+=Min;
top=inser;
u=e[s[top]^].to;
continue;
}
bool flag=false;
int v;
for(int i=cur[u];~i;i=e[i].next){
v=e[i].to;
if(e[i].cap-e[i].flow&&dep[v]+==dep[u]){
flag=true;
cur[u]=i;
break;
}
}
if(flag){
s[top++]=cur[u];
u=v;
continue;
}
int Min=n;
for(int i=head[u];~i;i=e[i].next)
if(e[i].cap-e[i].flow &&dep[e[i].to]<Min){
Min=dep[e[i].to];
cur[u]=i;
}
gap[dep[u]]--;
if(!gap[dep[u]])return ans;
gap[dep[u]=Min+]++;
if(u!=st)u=e[s[--top]^].to;
}
return ans;
}
int main(){
int t;
scanf("%d",&t);
for(int cas=;cas<=t;cas++){
printf("Case #%d: ",cas);
int n,m;
scanf("%d%d",&n,&m);
int in[N];
int st=,ed=n+;
memset(in,,sizeof in);
init();
for(int i=;i<=m;i++){
int u,v,d,b;
scanf("%d%d%d%d",&u,&v,&d,&b);
add(u,v,b);
in[v]+=d;
in[u]-=d;
}
int need=;
for(int i=;i<=n;i++){
if(in[i]>){
add(st,i,in[i]);
need+=in[i];
}
else add(i,ed,-in[i]);
}
int ans=sap(st, ed, ed+);
if(need==ans)puts("happy");
else puts("unhappy");
}
}