zoj2314 Reactor Cooling --- 上下界可行流

題目給出了每條邊的上下界，

此類題目的建邊方法是：

1、添加源點彙點，

2、對每條邊 添加邊 c(u,v) = up(u,v) - low(u,v)

3、對每個點 c(s,v) = out(v)

                       c(v,t) = in(v)   (權值為正)

求s到t的最大流，若最大流等于所有邊下界的和，則存在可行流，

每條邊的流量為 flow(u,v) +low(u,v)

#include <iostream>
#include <cstring>
#include <string>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <vector>
#include <queue>
#include <map>
#define inf 0x3f3f3f3f
#define eps 1e-6
#define ll __int64
const int maxn=210;
const int maxm=40010;
using namespace std;

int n,m,s,t,low[maxm],in[maxn],out[maxn];
struct node
{
    int from,to,cap,flow;
};
struct dinic
{
    int n,m,s,t;
    vector<node> e;
    vector<int> g[maxn];
    bool vis[maxn];
    int d[maxn];
    int cur[maxn];

    void init(int n)
    {
        e.clear();
        for(int i=0;i<=n+1;i++)
            g[i].clear();
    }

    void addedge(int a,int b,int c,int d)//c d為正向弧和反向弧的權值 一般反向弧為0
    {
        e.push_back((node){a,b,c,0});
        e.push_back((node){b,a,d,0});
        m=e.size();
        g[a].push_back(m-2);
        g[b].push_back(m-1);
    }

    bool bfs()
    {
        memset(vis,0,sizeof vis);
        queue<int> q;
        q.push(s);
        d[s]=0;
        vis[s]=1;
        while(!q.empty())
        {
            int x=q.front();q.pop();
            for(int i=0;i<g[x].size();i++)
            {
                node& ee=e[g[x][i]];
                if(!vis[ee.to]&&ee.cap>ee.flow)
                {
                    vis[ee.to]=1;
                    d[ee.to]=d[x]+1;
                    q.push(ee.to);
                }
            }
        }
        return vis[t];
    }

    int dfs(int x,int a)
    {
        if(x==t||a==0) return a;
        int flow=0,f;
        for(int& i=cur[x];i<g[x].size();i++)
        {
            node& ee=e[g[x][i]];
            if(d[x]+1==d[ee.to]&&(f=dfs(ee.to,min(a,ee.cap-ee.flow)))>0)
            {
                ee.flow+=f;
                e[g[x][i]^1].flow-=f;
                flow+=f;
                a-=f;
                if(a==0) break;
            }
        }
        return flow;
    }

    int maxflow(int s,int t)
    {
        this->s=s;
        this->t=t;
        int flow=0;
        while(bfs())
        {
            memset(cur,0,sizeof cur);
            flow+=dfs(s,inf);
        }
        return flow;
    }

    void output(int cnt)
    {
        for(int i=0;i<cnt;i++)
            printf("%d\n",e[i*2].flow+low[i]);
    }
};
dinic solve;

int main()
{
    int icy,a,b,cc,d,i,full;
    scanf("%d",&icy);
    while(icy--)
    {
        scanf("%d%d",&n,&m);
        s=0,t=n+1;
        solve.init(n);
        memset(in,0,sizeof in);
        memset(out,0,sizeof out);
        for(i=0;i<m;i++)
        {
            scanf("%d%d%d%d",&a,&b,&low[i],&d);
            solve.addedge(a,b,d-low[i],0);
            out[a]+=low[i];
            in[b]+=low[i];
        }
        full=0;
        for(i=1;i<=n;i++)
        {
            if(out[i]<in[i])
            {
                solve.addedge(s,i,in[i]-out[i],0);
                full+=in[i]-out[i];
            }
            else
                solve.addedge(i,t,out[i]-in[i],0);
        }
        if(solve.maxflow(s,t)!=full)
            printf("NO\n");
        else
        {
            printf("YES\n");
            solve.output(m);
        }
    }
    return 0;
}