題目連結：http://acm.hdu.edu.cn/showproblem.php?pid=5988

思路：求網路最小破壞可能，就是求最大不破壞的可能，再加上一個負號又變成了求最小費用流，對於乘法加上一個對數就變成了加法，需要注意的是對於浮點數要加上eps。

#pragma GCC optimize(2)
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <bitset>
#include <cmath>
#include <cctype>
#include <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <queue>
#include <map>
#include <set>
#include <sstream>
#include <iomanip>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const ll inff = 0x3f3f3f3f3f3f3f3f;
#define FOR(i,a,b) for(int i(a);i<=(b);++i)
#define FOL(i,a,b) for(int i(a);i>=(b);--i)
#define REW(a,b) memset(a,b,sizeof(a))
#define inf int(0x3f3f3f3f)
#define si(a) scanf("%d",&a)
#define sl(a) scanf("%lld",&a)
#define sd(a) scanf("%lf",&a)
#define ss(a) scanf("%s",a)
#define mod ll(6666666)
#define pb push_back
#define eps 1e-6
#define lc d<<1
#define rc d<<1|1
#define Pll pair<ll,ll>
#define P pair<int,int>
#define pi acos(-1)
const int N=1e2+7;
struct edge{int to,cap;
double cost;
int rev;};
vector<edge>g[N];
int pre[N],prv[N],n,m,s,t,flow;
double d[N],h[N],res;
void add(int from,int to,int cap,double cost)
{
    g[from].pb(edge{to,cap,cost,(int)g[to].size()});
    g[to].pb(edge{from,0,-cost,(int)g[from].size()-1});
}
void slove(int s,int t,int f,int n)
{
    //int res=0;
    fill(h,h+n+1,0);
    while(f>0)
    {
        priority_queue<pair<double,int>,vector<pair<double,int>>,greater<pair<double,int>> > q;
        fill(d,d+n+1,inf);
        d[s]=0;
        q.push(pair<double,int>(0,s));
        while(!q.empty())
        {
            P p=q.top();q.pop();
            int v=p.second;
            if(d[v]<p.first) continue;
            for(int i=0;i<(int)g[v].size();i++)
            {
                edge &e=g[v][i];
                if(e.cap>0&&d[e.to]>d[v]+e.cost+h[v]-h[e.to]+eps)
                {
                    d[e.to]=d[v]+e.cost+h[v]-h[e.to];
                    prv[e.to]=v;pre[e.to]=i;
                    q.push(pair<double,int>(d[e.to],e.to));
                }
            }
        }
        if(d[t]==inf) break;
        FOR(i,0,n) h[i]+=d[i];
        int d=f;
        for(int i=t;i!=s;i=prv[i])  d=min(d,g[prv[i]][pre[i]].cap);
        f-=d,flow+=d;
        res+=d*h[t];
        for(int i=t;i!=s;i=prv[i])
        {
            edge &e=g[prv[i]][pre[i]];
            e.cap-=d;
            g[i][e.rev].cap+=d;
        }
    }
    //return res;
}
int main()
{
    cin.tie(0);
    cout.tie(0);
    cin>>t;
    int x,y,z;
    double q;
    while(t--)
    {
        si(n),si(m);
        s=0;res=0.0;
        FOR(i,0,n+1) g[i].clear();
        FOR(i,1,n)
        {
            si(x),si(y);
            x-=y;
            if(x<0) add(i,n+1,abs(x),0);
            else if(x>0) add(s,i,x,0);
        }
        FOR(i,1,m)
        {
            si(x),si(y),si(z),sd(q);
            add(x,y,1,0);
            if(z>1) add(x,y,z-1,-log(1.0-q));
        }
        slove(s,n+1,inf,n+1);
        //cout<<flow<<endl;
        res=1.0-exp(-res);
        printf("%.2f\n",res);
    }
    return 0;
}