sgu 194上下界网络流

又搞了个模板，这个模板应该ok,足以应付各种网络流了

题意：给n个点m 条边，其中每条边的流量有两个限制不能大于r不能小于l,求是否有可行解，如有输出每条边的流量

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn = 1000000;
const int oo = 0x3777777;
struct arc{
    int to,cap,flow,next;
    arc(int x = 0,int y=0,int z = 0,int u =0){
        to = x;cap = y;
        flow = z;next = u;
    }
};
arc e[maxn];
int head[maxn],d[maxn],tot,id[maxn],work[maxn];
int S,T,q[maxn],hd,tail,n,m;
void add(int u,int v,int cap,int flow,int ID){
    id[tot] = ID;
    e[tot] = arc(v,cap,flow,head[u]);
    head[u] = tot++;
    id[tot] = 0;
    e[tot] = arc(u,0,-flow,head[v]);
    head[v] = tot++;
}

bool bfs(){
    for(int i=1;i<=T;i++)
    d[i] = 0;
    d[S] = 1;
    hd = tail = 0;
    q[tail++] = S;
    while(hd != tail){
        int u = q[hd++];
        if(hd == maxn)hd  =0;
        for(int i=head[u];i!=-1;i=e[i].next){
             if(e[i].cap > e[i].flow && !d[e[i].to]){
                d[e[i].to] = d[u] + 1;
                q[tail++] = e[i].to;
                if(tail == maxn)tail = 0;
                if(T == e[i].to)return true;
             }
        }
    }
    return false;
}
int dfs(int u,int low){
    int tmp = 0,f;
    if(u==T||!low)return low;
    for(int &i=work[u];i!=-1;i=e[i].next){
            if(d[e[i].to] == d[u] + 1 && (f=dfs(e[i].to,min(low,e[i].cap - e[i].flow)))){
               // tmp += f;
                e[i].flow += f;
                e[i^1].flow -= f;
              //  low -= f;
              //  if(!low)break;
              return f;
            }
    }
    return 0;
}
void init(){
    memset(head,-1,sizeof(head));
    tot = 0;
}
int du[maxn],dn[maxn],ans[maxn];
int main(){
    int u,v,down,up;
    while(~scanf("%d%d",&n,&m)){
        init();
        for(int i=1;i<=m;i++){
            scanf("%d%d%d%d",&u,&v,&down,&up);
            add(u,v,up-down,0,i);
            du[u] -= down;
            du[v] += down;
            dn[i] = down;
        }
        S = 0;T=n+1;
        int Edge = tot;
        for(int i=1;i<=n;i++){
            if(du[i] > 0) add(S,i,du[i],0,0);
            if(du[i] < 0) add(i,T,-du[i],0,0);
        }
        while(bfs())
        {
            for(int i=0; i<n+2; i++)
            work[i]=head[i];
            while(dfs(S,oo));
        }
        int flag = 0;
        for(int i=head[S];i!=-1;i=e[i].next){
            if(e[i].flow != e[i].cap){
                flag = 1;break;
            }
        }
        if(flag)puts("NO");
        else {
            puts("YES");
            for(int i = 0;i<Edge;i++) ans[id[i]] = e[i].flow;
            for(int i=1;i<=m;i++)
                printf("%d\n",ans[i]+dn[i]);
        }

    }
}