Harmonious Army

Harmonious Army

Now, Bob is playing an interesting game in which he is a general of a harmonious army. There are n soldiers in this army. Each soldier should be in one of the two occupations, Mage or Warrior. There are m pairs of soldiers having combination ability. There are three kinds of combination ability. If the two soldiers in a pair are both Warriors, the army power would be increased by a. If the two soldiers in a pair are both Mages, the army power would be increased by c. Otherwise the army power would be increased by b, and b=a/4+c/3, guaranteed that 4|a and 3|c. Your task is to output the maximum power Bob can increase by arranging the soldiers' occupations.

Note that the symbol a|b means that a divides b, e.g. , 3|12 and 8|24.

Input

There are multiple test cases.

Each case starts with a line containing two positive integers n(n≤500) and m(m≤\(10^4\)).

In the following m lines, each line contains five positive integersu,v,a,b,c (1≤u,v≤n,u≠v,1≤a,c≤4×\(10^6\),b=a/4+c/3), denoting soldiers u and v have combination ability, guaranteed that the pair (u,v) would not appear more than once.

It is guaranteed that the sum of n in all test cases is no larger than 5×\(10^3\), and the sum of m in all test cases is no larger than 5×\(10^4\).

Output

For each test case, output one line containing the maximum power Bob can increase by arranging the soldiers' occupations.

Sample Input

3 2

1 2 8 3 3

2 3 4 3 6

Sample Output

12

思路：

把\(s\)连向\(u，v\)连容量为\(\frac{a+b}{2}\)的有向边\(,\)把\(u,v\)向\(t\)连容量为\(\frac{b+c}{2}\)的有向边\(，\)在\(u,v\)间连条\(\frac{a+c}{2}-b\)，用\(\sum_{i=1}^n(a_i+b_i+c_i)\)减最大流
如图：

证明：


(这张图找了半天)

\[\begin{cases} a+b=A+B&\\ c+d=B+C&\\ a+e+d=b+e+c=A+C \end{cases}\]

其中一组解为

\[\begin{cases} a=b=\frac{A+B}{2}&\\ c=d=\frac{B+C}{2}&\\ e=\frac{A+C}{2}-B \end{cases}\]

便得到了上面建边的由来

\(\mathfrak{Talk\ is\ cheap,show\ you\ the\ code.}\)

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
# define read read1<int>()
# define Type template<typename T>
Type inline T read1(){
    T n=0;
    char k;
    bool fl=0;
    do (k=getchar())=='-'&&(fl=1);while('9'<k||k<'0');
    while(47<k&&k<58)n=(n<<3)+(n<<1)+(k^48),k=getchar();
    return fl?-n:n;
}
class Flow{
	# define Big 502
	# define Big2 70000
    # define Imax 2e9
	private:
		int cur[Big+1],q[Big+1],lev[Big+1];
		class Pic{
			public:
				class Edge{
					public: 
						int u,t;
                        double v;
				}G[Big2<<1|1];
				int f[Big+1],tot;
				Pic(){for(int i=0;i<=Big;++i)f[i]=0;tot=0;}
				void add(int u,int v,double t){
					G[++tot].u=v;
					G[tot].t=f[u];
					f[u]=tot;
					G[tot].v=t;
				}
		}G;
	public:
		int S,T,n;
        void Clear(){memset(G.f,0,sizeof(G.f));G.tot=0;}
        Flow(){n=Big;}
		Flow(int big,int s,int t){n=big;S=s;T=t;}
        void Into(int s,int t){S=s;T=t;}
		void add(int u,int v,double t){
			G.add(u,v,t);G.add(v,u,0);
		}
		bool bfs(){
			for(int i=0;i++^n;lev[i]=0);
			lev[S]=1;int t=0;
			q[t++]=S;
			for(int i=0;i^t;++i){
				int u=q[i];
                //printf("->%d\n",u);
				for(int i=G.f[u];i;i=G.G[i].t){
					if(!lev[G.G[i].u]&&G.G[i].v>0){
						int to=G.G[i].u;
						lev[to]=lev[u]+1;
						q[t++]=to;
						if(to==T)return 1;
					}
				}
			}
			return 0;
		}
		double dfs(int u,double f){
			if(u==T)return f;
			double tag=0,c;
			for(int &i=cur[u];i;i=G.G[i].t){
				int to=G.G[i].u;
				if(G.G[i].v>0&&lev[to]==lev[u]+1){
					c=dfs(to,min(f-tag,G.G[i].v));
					G.G[i].v-=c;
					G.G[(i-1^1)+1].v+=c;
					tag+=c;
					if(tag==f)return tag; 
				}
			}
			return tag;
		}
		double Dinic(){
			double ans=0;
			while(bfs()){
				for(int i=0;i++^n;)cur[i]=G.f[i];
				ans+=dfs(S,2e9);
			}
			return ans;
		}
	# undef Big
	# undef Big2
}G;
# define fre(k) freopen(k".in","r",stdin);freopen(k".out","w",stdout)
# define f(i,l,r) for(int i=l;i<=r;++i)
int main(){
    G.Into(1,2);
    int n,m;
    while(~scanf("%d %d",&n,&m)){
        double t=0;
        G.Clear();
        for(int i=0;i++^m;){
            int u=read,v=read;
            double A=read,B=read,C=read;
            t+=A+B+C;
            G.add(1,u+2,(A+B)/2);
            G.add(1,v+2,(A+B)/2);
            G.add(u+2,2,(B+C)/2);
            G.add(v+2,2,(B+C)/2);
            G.add(u+2,v+2,(A+C)/2-B);
            G.add(v+2,u+2,(A+C)/2-B);
        }
        printf("%.0f\n",t-G.Dinic());
    }
    return 0;
}