抽象至极的一道题。 第一问想到用网络流；第二问拆分函数并最大流求，分得越细，即△的值越小，精度就越优秀，但复杂度就越劣；合理控制△的值，可以拿到一定的分数。

#include<bits/stdc++.h>
#define x first
#define y second
using namespace std;
typedef long long LL;
const double eps=1e-8;
const int N=1100,M=1100,INF=0x3f3f3f3f;
inline int read(){
	char c=getchar();int x=0,f=1;
	while(c<'0'||'9'<c){if(c=='-')f=-1;c=getchar();}
	while('0'<=c&&c<='9'){x=x*10+c-'0';c=getchar();}
	return x*f;
}

int cnt=1,head[N+M];
struct edge{int from,to,next; double val;}e[N*M<<1];
inline void add(int f,int t,double v)
	{e[++cnt]={f,t,head[f],v};head[f]=cnt;/*if(v)printf("%d %d %.1lf\n",f,t,(v && v<=10)? v:0);*/}
inline LL gcd(LL a,LL b)
	{while(b){a%=b; a^=b; b^=a; a^=b;}return a;}
inline void initiator()
	{memset(head,0,sizeof head); cnt=1;}
struct frac		//val=x/y
{
	LL x,y;
	frac(){x=0;	y=1;}
	frac(LL a,LL b=1):x(a),y(b)
	{
		LL g=gcd(a,b);
		x/=g;	y/=g;
	}
	inline double value(){return (double)x/y;}
};
typedef pair<frac,frac> point;
struct line{frac k,b;};			//y=kx+b
inline frac simplify(frac a)
{
	if(!a.x) return frac(0,1);
	if(a.y<0)	a.x=-a.x,a.y=-a.y;
	LL g=gcd(abs(a.x),a.y);
}
inline frac simplify(LL a,LL b){return simplify(frac(a,b));}
inline frac operator+(frac a,frac b){return simplify(a.x*b.y+a.y*b.x,a.y*b.y);}
inline frac operator-(frac a,frac b){return simplify(a.x*b.y-a.y*b.x,a.y*b.y);}
inline frac operator*(frac a,frac b){return simplify(a.x*b.x,a.y*b.y);}
inline frac operator/(frac a,frac b){return simplify(a.x*b.y,a.y*b.x);}
inline bool operator==(frac a,frac b){return a.x==b.x && a.y==b.y;}
inline bool operator==(line a,line b){return a.k==b.k && a.b==b.b;}
inline void print(frac a){printf("%lld/%lld\n",a.x,a.y);}
int n,m,a[N],b[N],c[N],d[M],dep[N+M],now[N+M],s=0,t;
bool graph[N][M];

inline bool bfs()
{
	memset(dep,0,sizeof dep);
	queue<int> q;	dep[s]=1;
	q.push(s);	now[s]=head[s];
	while(!q.empty())
	{
		int u=q.front();	q.pop();
		for(int i=head[u];i;i=e[i].next)
			if(e[i].val>0)
			{
				int v=e[i].to;
				if(dep[v])	continue;
				dep[v]=dep[u]+1;
				now[v]=head[v];
				if(v==t) return 1;
				q.push(v);
			}
	}
}

double dinic(int u,double flow)
{
	if(u==t)	return flow;
	double last=flow;
	for(int &i=now[u];i;i=e[i].next)
		if(e[i].val>0)
		{
			int v=e[i].to;
			if(dep[v]!=dep[u]+1)	continue;
			double k=dinic(v,min(last,e[i].val));
			if(!k)		dep[v]=-1;
			last-=k;	e[i].val-=k;
			e[i^1].val+=k;
			if(!last)	break;
		}
}

line solve(double x)
{
	initiator();
	for(int i=1;i<=n;++i)
		if(b[i]<x)
		{
			double lim=a[i]? min((x-b[i])/2/a[i],(double)c[i]):c[i];
			add(s,i,lim);	add(i,s,0);
		}
	for(int i=1;i<=n;++i)
		for(int j=1;j<=m;++j)
			if(graph[i][j])
				add(i,n+j,1e18),add(n+j,i,0);
	for(int i=1;i<=m;++i)
		add(n+i,t,d[i]),add(t,n+i,0);
	while(bfs()) dinic(s,1e18);
	line res;
	for(int i=2;i<=cnt;i+=2)
		if(!dep[e[i].to])				//注意这里不能直接判流量（double 永远的痛）
		{
			if(e[i].from==s)
			{
				int u=e[i].to;
				if(a[u] && x<2*a[u]*c[u]+b[u])
				{
					res.k=res.k+(frac)1/2/a[u];
					res.b=res.b-(frac)b[u]/2/a[u];
//					printf("for edge %d->%d k+=%d/%d\n",s,e[i].to,((frac)1/2/a[u]).x,((frac)1/2/a[u]).y);
				}
				else res.b=res.b+c[u];//,printf("for edge %d->%d b+=%d\n",s,e[i].to,c[u]);
			}
			else if(e[i].to==t && dep[e[i].from])	//说明这条边是流量限制
				res.b=res.b+d[e[i].from-n];//,printf("for edge %d->%d b+=%d\n",e[i].from,t,d[e[i].from-n]);
		}
//	printf("res.k=%d/%d\n",res.k.x,res.k.y);
}

vector<point> v;
void solve(line l,line r)
{
//	printf("solving (%d/%d)x+(%d/%d) ~ (%d/%d)x+(%d/%d)\n",l.k.x,l.k.y,l.b.x,l.b.y,r.k.x,r.k.y,r.b.x,r.b.y);
	if(l.k==r.k)	return;
	frac x=(r.b-l.b)/(l.k-r.k);
	point p=make_pair(x,x*l.k+l.b);
	line midl=solve(p.x.value()-eps);
	line midr=solve(p.x.value()+eps);
	if(midr==r)	{/*printf("pushing (%d/%d,%d/%d)\n",p.x.x,p.x.y,p.y.x,p.y.y);*/	v.push_back(p); return;}
	else	solve(l,midl),solve(midr,r);
}

int main()
{
//	freopen("mincost10.in","r",stdin);
//	freopen("mincost10.out","w",stdout);
	n=read();	m=read();	t=n+m+1;
	for(int i=1;i<=n;++i)
		a[i]=read(),b[i]=read(),c[i]=read();
	for(int i=1;i<=m;++i)	d[i]=read();
	for(int i=1;i<=n;++i)
		for(int j=1;j<=m;++j)
			graph[i][j]=read();
	line r=solve(1e9);	frac ans;	assert(r.b.y==1);
	v.push_back(make_pair(frac(),frac()));
	for(int i=1;i<=3;++i)
	{
		line left=solve(i-1+eps);
		line right=solve(i-eps);
		v.push_back(make_pair(frac(i-1),left.k*(i-1)+left.b));
		solve(left,right);
		v.push_back(make_pair(frac(i),right.k*i+right.b));
	}
	line now=solve(3+eps);
	v.push_back(make_pair(frac(3),now.k*3+now.b));
	solve(solve(3+eps),r);
	for(unsigned i=1;i<v.size();++i)
	{
		frac now=(v[i].x+v[i-1].x)*(v[i].y-v[i-1].y)/2;
//		printf("now ans+=%d/%d\n",now.x,now.y);
		ans=ans+now;
	}
}