SGU 194 Reactor Cooling ——网络流

【题目分析】

    无源汇上下界可行流。

    上下界网络流的问题可以参考这里。↓

    http://www.cnblogs.com/kane0526/archive/2013/04/05/3001108.html

【代码】

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>

//#include <map>
#include <set>
#include <queue>
#include <string>
#include <iostream>
#include <algorithm>

using namespace std;

#define maxn 205
#define me 50005
#define inf 0x3f3f3f3f
#define F(i,j,k) for (int i=j;i<=k;++i)
#define D(i,j,k) for (int i=j;i>=k;--i)

void Finout()
{
    #ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
//    freopen("out.txt","w",stdout);
    #endif
}

int Getint()
{
    int x=0,f=1; char ch=getchar();
    while (ch<'0'||ch>'9') {if (ch=='-') f=-1; ch=getchar();}
    while (ch>='0'&&ch<='9') {x=x*10+ch-'0'; ch=getchar();}
    return x*f;
}

int h[me<<1],to[me<<1],ne[me<<1],fl[me<<1],en=0,S=0,T=me-1;
int id[me<<1];
 
void add(int a,int b,int c,int ID)
{
//	cout<<"add "<<a<<" "<<b<<" "<<c<<endl;
    to[en]=b; ne[en]=h[a]; fl[en]=c; id[en]=ID; h[a]=en++;
    to[en]=a; ne[en]=h[b]; fl[en]=0; id[en]=0; h[b]=en++;
}
 
int map[me];
 
bool tell()
{
    queue <int> q;
    memset(map,-1,sizeof map);
    map[S]=0;
    while (!q.empty()) q.pop();
    q.push(S);
    while (!q.empty())
    {
        int x=q.front(); q.pop();
//        cout<<"bfs"<<x<<endl;
        for (int i=h[x];i>=0;i=ne[i])
        {
//        	cout<<"to "<<to[i]<<endl;
            if (map[to[i]]==-1&&fl[i]>0)
            {
                map[to[i]]=map[x]+1;
                q.push(to[i]);
            }
        }
    }
//    cout<<"over"<<endl;
    if (map[T]!=-1) return true;
    return false;
}
 
int zeng(int k,int r)
{
    if (k==T) return r;
    int ret=0;
    for (int i=h[k];i>=0&&ret<r;i=ne[i])
        if (map[to[i]]==map[k]+1&&fl[i]>0)
        {
            int tmp=zeng(to[i],min(fl[i],r-ret));
            ret+=tmp; fl[i]-=tmp; fl[i^1]+=tmp;
        }
    if (!ret) map[k]=-1;
    return ret;
}

int ans[me<<1],n,m,du[me<<1],dn[me<<1];

int main()
{
    Finout();
    while (scanf("%d%d",&n,&m)!=EOF)
    {
    	memset(h,-1,sizeof h);
    	memset(ans,0,sizeof ans);
    	memset(du,0,sizeof du);
    	memset(dn,0,sizeof dn);
    	F(i,1,m)
    	{
    		int a,b,c,d;
    		a=Getint();b=Getint();c=Getint();d=Getint();
    		add(a,b,d-c,i);
    		du[a]-=c; du[b]+=c; dn[i]+=c;
		}
		F(i,1,n)
		{
			if (du[i]) add(S,i,du[i],0);
			if (du[i]<0) add(i,T,-du[i],0);
		}
		int now=0,tmp=0;
		while (tell()) while (tmp=zeng(S,inf)) now+=tmp;
		int flag=1;
		for (int i=h[S];i>=0;i=ne[i])
			if (fl[i]>0) flag=0;
		if (!flag) printf("NO
");
		else
		{
			printf("YES
");
			F(i,0,en-1) ans[id[i]]=fl[i^1];
			F(i,1,m) printf("%d
",ans[i]+dn[i]);
		}
	}
}