SGU 176 Flow construction 有源汇上下界网络流 最小流

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;}