【有上下界网络流】【ZOJ】2314 Reactor Cooling

【算法】有上下界网络流-无源汇(循环流)

【题解】http://www.cnblogs.com/onioncyc/p/6496532.html

//未提交
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
const int maxn=,maxm=,inf=0x3f3f3f3f;
struct edge{int from,v,flow;}e[maxm];
int S,T,TT,n,m,tot=,first[maxn],cur[maxn],in[maxn],d[maxn],q[],lows[maxm];
void insert(int u,int v,int w)
{
    tot++;e[tot].v=v;e[tot].flow=w;e[tot].from=first[u];first[u]=tot;
    tot++;e[tot].v=u;e[tot].flow=;e[tot].from=first[v];first[v]=tot;
}
bool bfs()
{
    memset(d,-,sizeof(d));
    int head=,tail=;q[]=S;d[S]=;
    while(head!=tail)
     {
         int x=q[head++];if(head>)head=;
         for(int i=first[x];i;i=e[i].from)
          if(d[e[i].v]==-&&e[i].flow)
           {
               d[e[i].v]=d[x]+;
               q[tail++]=e[i].v;
               if(tail>)tail=;
          }
     }
    return d[T]!=-;
}
int dfs(int x,int a)
{
    if(x==T||a==)return a;
    int flow=,f;
    for(int& i=cur[x];i;i=e[i].from)
     if(d[e[i].v]==d[x]+&&(f=dfs(e[i].v,min(a,e[i].flow)))>)
      {
          e[i].flow-=f;
          e[i^].flow+=f;
          a-=f;
          flow+=f;
          if(a==)break;
      }
    return flow;
}
int main()
{
    scanf("%d",&TT);
    while(TT--)
     {
         memset(first,,sizeof(first));
         memset(in,,sizeof(in));
         tot=;
         scanf("%d%d",&n,&m);
         for(int i=;i<=m;i++)
          {
              int u,v,w;
              scanf("%d%d%d%d",&u,&v,&lows[i],&w);
              in[u]-=lows[i];in[v]+=lows[i];
              insert(u,v,w-lows[i]);
         }
        S=,T=n+;
        for(int i=;i<=n;i++)
          {
             if(in[i]>)insert(S,i,in[i]);
             if(in[i]<)insert(i,T,-in[i]);
          }
         while(bfs())
          {
              for(int i=;i<=n+;i++)cur[i]=first[i];
            dfs(S,inf);
         }
        bool ok=;
        for(int i=first[S];i;i=e[i].from)
         if(e[i].flow)ok=;
        if(ok)
         {
             printf("YES\n");
             for(int i=;i<=m*;i+=)
              printf("%d\n",e[i^].flow+lows[i>>]);
         }
        else printf("NO\n");
     }
    return ;
}