LOJ.115.[模板]无源汇有上下界可行流(Dinic)
题目链接
参考:http://blog.csdn.net/clove_unique/article/details/54884437
http://blog.csdn.net/wu_tongtong/article/details/73320968
//输出一条边的流量 即输出它反向边上的流量与下界之和
#include<cstdio>
#include<cctype>
#include<algorithm>
#define gc() getchar()
typedef long long LL;
const int N=205,M=10205+N;
const LL INF=1e12;
int n,m,srcc,dess,lev[N],q[N],Enum,cur[N],H[N],nxt[M<<1],to[M<<1];
LL dgr[N],cap[M<<1],low[M],upp;
inline LL read()
{
LL now=0;register char c=gc();
for(;!isdigit(c);c=gc());
for(;isdigit(c);now=now*10+c-'0',c=gc());
return now;
}
inline void AddEdge(int u,int v,LL w)
{
to[++Enum]=v, nxt[Enum]=H[u], cap[Enum]=w, H[u]=Enum;
to[++Enum]=u, nxt[Enum]=H[v], cap[Enum]=0, H[v]=Enum;
}
bool BFS()
{
for(int i=srcc;i<=dess;++i) cur[i]=H[i],lev[i]=0;
lev[srcc]=1, q[0]=srcc;
int h=0,t=1;
while(h<t)
{
int x=q[h++];
for(int i=H[x];i;i=nxt[i])
if(!lev[to[i]] && cap[i])
{
lev[to[i]]=lev[x]+1,q[t++]=to[i];
if(to[i]==dess) return 1;
}
}
return 0;
}
LL Dinic(int u,LL flow)
{
if(u==dess) return flow;
LL used=0,delta;
for(int &i=cur[u];i;i=nxt[i])
if(lev[to[i]]==lev[u]+1 && cap[i])
{
delta=Dinic(to[i],std::min(flow-used,cap[i]));
if(delta)
{
cap[i]-=delta, cap[i^1]+=delta, used+=delta;
if(used==flow) return flow;
}
}
lev[u]=0;
return used;
}
int main()
{
n=read(),m=read();
Enum=1, srcc=0, dess=n+1;
LL tmp=0;
for(int u,v,i=1;i<=m;++i)
{
u=read(),v=read(),low[i]=read(),upp=read(),
dgr[u]-=low[i], dgr[v]+=low[i], AddEdge(u,v,upp-low[i]);
}
for(int i=1;i<=n;++i)
if(dgr[i]>0) AddEdge(srcc,i,dgr[i]),dgr[srcc]+=dgr[i];
else if(dgr[i]<0) AddEdge(i,dess,-dgr[i]);
while(BFS()) tmp+=Dinic(srcc,INF);
if(tmp==dgr[srcc])
{
puts("YES");
for(int i=2;i<=m<<1;i+=2) printf("%lld\n",cap[i^1]+low[i>>1]);
}
else puts("NO");
return 0;
}
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很久以前的奇怪但现在依旧成立的签名
attack is our red sun $$\color{red}{\boxed{\color{red}{attack\ is\ our\ red\ sun}}}$$ ------------------------------------------------------------------------------------------------------------------------
很久以前的奇怪但现在依旧成立的签名
attack is our red sun $$\color{red}{\boxed{\color{red}{attack\ is\ our\ red\ sun}}}$$ ------------------------------------------------------------------------------------------------------------------------