题目大意

给出N个点，M条有向边
如果有向边的标号是1的话，就表示该边的上界下界都为容量
如果有向边的标号为0的哈，表示该边的下界为0，上界为容量
现在问，从1到N的最小流是多少，并输出每条边的流量

分析

像无源汇点上下界可行流那样建图，然后从SS->ST跑一次，然后t->s连一条容量为+∞的边，跑一次，t->s这条边的流量即为最小流。

代码

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
#define INF 0x7f7f7f7f
#define MAXN 100
using namespace std;
int n,m,d[MAXN+10],S,T,num,dist[MAXN+10],vd[MAXN+10],tot,l[MAXN*MAXN+10],flow;
void Read(int &x){char c;while(c=getchar(),c!=EOF)if(c>='0'&&c<='9'){x=c-'0';while(c=getchar(),c>='0'&&c<='9')x=x*10+c-'0';ungetc(c,stdin);return;}
}
struct node{int v,cap;node *next,*back;
}*adj[MAXN+10],edge[MAXN*MAXN+10],*ecnt=edge,*epos[MAXN*MAXN+10];
void addedge(int u,int v,int cap,int pos){node *p=++ecnt;p->v=v;p->cap=cap;p->next=adj[u];epos[pos]=p;adj[u]=p;p=p->back=++ecnt;p->v=u;p->cap=0;p->next=adj[v];adj[v]=p;p->back=ecnt-1;
}
void read(){Read(n),Read(m);int i,u,v,z,c;for(i=1;i<=m;i++){Read(u),Read(v),Read(z),Read(c);if(c==1)d[u]-=z,d[v]+=z,epos[i]=0,l[i]=z;elseaddedge(u,v,z,i);}S=n+1,T=n+2;for(i=1;i<=n;i++)if(d[i]>0)addedge(S,i,d[i],0),tot+=d[i];else if(d[i]<0)addedge(i,T,-d[i],0);
}
int dfs(int u,int augu){if(u==T)return augu;int v,augv=0,delta,mind=num-1;for(node *p=adj[u];p;p=p->next)if(p->cap){v=p->v;if(dist[u]==dist[v]+1){delta=min(augu-augv,p->cap);delta=dfs(v,delta);augv+=delta;p->cap-=delta;p->back->cap+=delta;if(augu==augv||dist[S]>=num)return augv;}mind=min(mind,dist[v]);}if(!augv){if(!--vd[dist[u]])dist[S]=num;dist[u]=mind+1;vd[dist[u]]++;}return augv;
}
void mcmf(){memset(dist,0,sizeof dist);memset(vd,0,sizeof vd);num=n+2;vd[0]=num;while(dist[S]<num)flow+=dfs(S,INF);
}
void print(){if(flow!=tot){puts("Impossible");return;}printf("%d\n",epos[m+1]->back->cap);for(int i=1;i<m;i++)if(epos[i])printf("%d ",epos[i]->back->cap);elseprintf("%d ",l[i]);if(epos[m])printf("%d\n",epos[m]->back->cap);elseprintf("%d\n",l[m]);
}
int main()
{read();mcmf();addedge(n,1,INF,m+1);mcmf();print();
}