题意：有n个节点m根水管的网络，每根水管有个水流的限制，问最少要多少水流能满足整个网络。

最小流
代码：

//author: CHC
//First Edit Time:	2015-09-10 10:45
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <set>
#include <vector>
#include <map>
#include <queue>
#include <set>
#include <algorithm>
#include <limits>
using namespace std;
typedef long long LL;
const int MAXN=110;
const int MAXM=1e+5;
const int INF = numeric_limits<int>::max();
const LL LL_INF= numeric_limits<LL>::max();
struct Edge {
    int to,ci,next;
    Edge(){}
    Edge(int _to,int _ci,int _next):to(_to),ci(_ci),next(_next){}
}e[MAXM];
int head[MAXN],tot;
void init(){
    memset(head,-1,sizeof(head));
    tot=0;
}
void AddEdge(int u,int v,int ci0,int ci1=0){
    e[tot]=Edge(v,ci0,head[u]);
    head[u]=tot++;
    e[tot]=Edge(u,ci1,head[v]);
    head[v]=tot++;
}
int dis[MAXN],sta[MAXN],top,cur[MAXN];
bool bfs(int st,int et){
    memset(dis,0,sizeof(dis));
    dis[st]=1;
    queue <int> q;q.push(st);
    while(!q.empty()){
        int now=q.front();q.pop();
        for(int i=head[now];~i;i=e[i].next)
            if(e[i].ci&&!dis[e[i].to]){
                dis[e[i].to]=dis[now]+1;
                q.push(e[i].to);
            }
    }
    return dis[et];
}
LL Dinic(int st,int et){
    LL ans=0;
    while(bfs(st,et)){
        top=0;
        memcpy(cur,head,sizeof(cur));
        int u=st,i;
        while(1){
            if(u==et){
                int pos,minn=INF;
                for(i=0;i<top;i++){
                    if(minn>e[sta[i]].ci){
                        minn=e[sta[i]].ci;
                        pos=i;
                    }
                }
                for(i=0;i<top;i++){
                    e[sta[i]].ci-=minn;
                    e[sta[i]^1].ci+=minn;
                }
                top=pos;
                u=e[sta[top]^1].to;
                ans+=minn;
            }
            for(i=cur[u];~i;cur[u]=i=e[i].next)
                if(e[i].ci&&dis[u]+1==dis[e[i].to])break;
            if(cur[u]!=-1){
                sta[top++]=cur[u];
                u=e[cur[u]].to;
            }
            else {
                if(top==0)break;
                dis[u]=0;
                u=e[sta[--top]^1].to;
            }
        }
    }
    return ans;
}
int du[MAXN],n,m,num[MAXM],base[MAXM];
int main()
{
    while(~scanf("%d%d",&n,&m)){
        memset(du,0,sizeof(du));
        init();
        for(int i=0,x,y,ci,v;i<m;i++){
            scanf("%d%d%d%d",&x,&y,&ci,&v);
            num[i]=tot;
            if(!v){
                AddEdge(x,y,ci);
                base[i]=0;
            }
            else {
                AddEdge(x,y,0);
                du[x]-=ci;du[y]+=ci;
                base[i]=ci;
            }
        }
        int sst=0,eet=n+1;
        LL ctot=0;
        for(int i=1;i<=n;i++)
            if(du[i]>0)AddEdge(sst,i,du[i]),ctot+=du[i];
            else AddEdge(i,eet,-du[i]);
        LL ans=Dinic(sst,eet);
        AddEdge(n,1,INF);
        //ans+=Dinic(sst,eet);
        LL t=Dinic(sst,eet);
        if(ans+t!=ctot)puts("Impossible");
        else {
            //ans=Dinic(1,n);
            //printf("%d\n",e[tot-1].ci);
            printf("%I64d\n",t);
            for(int i=0;i<m-1;i++){
                printf("%d ",base[i]+e[num[i]^1].ci);
            }
            printf("%d\n",base[m-1]+e[num[m-1]^1].ci);
        }
    }
    return 0;
}