【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=1000,maxm=100000,inf=0x3f3f3f3f;
struct edge{int from,v,flow;}e[maxm];
int S,T,TT,n,m,tot=1,first[maxn],cur[maxn],in[maxn],d[maxn],q[1010],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=0;e[tot].from=first[v];first[v]=tot;
}
bool bfs()
{
    memset(d,-1,sizeof(d));
    int head=0,tail=1;q[0]=S;d[S]=0;
    while(head!=tail)
     {
         int x=q[head++];if(head>1000)head=0;
         for(int i=first[x];i;i=e[i].from)
          if(d[e[i].v]==-1&&e[i].flow)
           {
               d[e[i].v]=d[x]+1;
               q[tail++]=e[i].v;
               if(tail>1000)tail=0;
          }
     }
    return d[T]!=-1;
}
int dfs(int x,int a)
{
    if(x==T||a==0)return a;
    int flow=0,f;
    for(int& i=cur[x];i;i=e[i].from)
     if(d[e[i].v]==d[x]+1&&(f=dfs(e[i].v,min(a,e[i].flow)))>0)
      {
          e[i].flow-=f;
          e[i^1].flow+=f;
          a-=f;
          flow+=f;
          if(a==0)break;
      }
    return flow;
}
int main()
{
    scanf("%d",&TT);
    while(TT--)
     {
         memset(first,0,sizeof(first));
         memset(in,0,sizeof(in));
         tot=1;
         scanf("%d%d",&n,&m);
         for(int i=1;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=0,T=n+1;
        for(int i=1;i<=n;i++)
          {
             if(in[i]>0)insert(S,i,in[i]);
             if(in[i]<0)insert(i,T,-in[i]);
          }
         while(bfs())
          {
              for(int i=0;i<=n+1;i++)cur[i]=first[i];
            dfs(S,inf);
         }
        bool ok=1;
        for(int i=first[S];i;i=e[i].from)
         if(e[i].flow)ok=0;
        if(ok)
         {
             printf("YES\n");
             for(int i=2;i<=m*2;i+=2)
              printf("%d\n",e[i^1].flow+lows[i>>1]);
         }
        else printf("NO\n");
     }
    return 0;
}