去年通话邀请赛的B题，当时居然过的那么少。。。明明是一道非常裸的可行流最小流麽。。仅仅要对每种人分别求一下可行最小流加起来就能够了。建图是对每一个点拆点，容量上下届都设为v[i]，然后每一个点间能连边的直接连边就能够了。然后在这个图的基础上转化为可行流最小流，求一下就能够了。。。
#include<algorithm>
#include<iostream>
#include<cstring>
#include<vector>
#include<cstdio>
#include<cmath>
#include<queue>
#include<stack>
#include<map>
#include<set>
#define LL long long
#define inf 0x3f3f3f3f
#define CLR(a, b) memset(a, b, sizeof(a))
using namespace std; const int maxn = 440;
const int INF = 0x3f3f3f3f; struct Edge
{
int from, to, cap, flow;
Edge() {}
Edge(int from, int to, int cap, int flow)
:from(from), to(to), cap(cap), flow(flow) {}
}; struct ISAP
{
int n, m, s, t;
vector<Edge> edges;
vector<int> G[maxn]; // 邻接表，G[i][j]表示结点i的第j条边在e数组中的序号
bool vis[maxn]; // BFS使用
int d[maxn]; // 从起点到i的距离
int cur[maxn]; // 当前弧指针
int p[maxn]; // 可增广路上的上一条弧
int num[maxn]; // 距离标号计数 void AddEdge(int from, int to, int cap)
{
edges.push_back(Edge(from, to, cap, 0));
edges.push_back(Edge(to, from, 0, 0));
m = edges.size();
G[from].push_back(m-2);
G[to].push_back(m-1);
} bool BFS()
{
memset(vis, 0, sizeof(vis));
queue<int> Q;
Q.push(t);
vis[t] = 1;
d[t] = 0;
while(!Q.empty())
{
int x = Q.front();
Q.pop();
for(int i = 0; i < G[x].size(); i++)
{
Edge& e = edges[G[x][i]^1];
if(!vis[e.from] && e.cap > e.flow)
{
vis[e.from] = 1;
d[e.from] = d[x] + 1;
Q.push(e.from);
}
}
}
return vis[s];
} void init(int n)
{
this->n = n;
for(int i = 0; i < n; i++) G[i].clear();
edges.clear();
} int Augment()
{
int x = t, a = INF;
while(x != s)
{
Edge& e = edges[p[x]];
a = min(a, e.cap-e.flow);
x = edges[p[x]].from;
}
x = t;
while(x != s)
{
edges[p[x]].flow += a;
edges[p[x]^1].flow -= a;
x = edges[p[x]].from;
}
return a;
} int Maxflow(int s, int t, int need)
{
this->s = s;
this->t = t;
int flow = 0;
BFS();
memset(num, 0, sizeof(num));
for(int i = 0; i < n; i++) num[d[i]]++;
int x = s;
memset(cur, 0, sizeof(cur));
while(d[s] < n)
{
if(x == t)
{
flow += Augment();
if(flow >= need) return flow;
x = s;
}
int ok = 0;
for(int i = cur[x]; i < G[x].size(); i++)
{
Edge& e = edges[G[x][i]];
if(e.cap > e.flow && d[x] == d[e.to] + 1) // Advance
{
ok = 1;
p[e.to] = G[x][i];
cur[x] = i; // 注意
x = e.to;
break;
}
}
if(!ok) // Retreat
{
int m = n-1; // 初值注意
for(int i = 0; i < G[x].size(); i++)
{
Edge& e = edges[G[x][i]];
if(e.cap > e.flow) m = min(m, d[e.to]);
}
if(--num[d[x]] == 0) break;
num[d[x] = m+1]++;
cur[x] = 0; // 注意
if(x != s) x = edges[p[x]].from;
}
}
return flow;
}
} sol; int n, m; struct Point
{
int x, y, b, e;
int v[7];
void inpt()
{
scanf("%d%d%d%d", &x, &y, &b, &e);
e += b;
for(int i = 0; i < m; i ++)
scanf("%d", &v[i]);
}
}p[maxn]; bool ok(Point a, Point b)
{
int len = (a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y);
return b.b > a.e && len <= (b.b - a.e) * (b.b - a.e);
} int solve(int idx)
{
sol.init(n * 2 + 5);
int S = 0, T = 2 * n + 1, SS = 2 * n + 2, ST = SS + 1;
for(int i = 1; i <= n; i ++)
{
// sol.AddEdge(i, i + n, 0);
sol.AddEdge(S, i, INF);
sol.AddEdge(i + n, T, INF); sol.AddEdge(SS, i + n, p[i].v[idx]);
sol.AddEdge(i, ST, p[i].v[idx]);
}
for(int i = 1; i <= n; i ++)
for(int j = 1; j <= n; j ++)
{
if(!ok(p[i], p[j])) continue;
sol.AddEdge(i + n, j, p[i].v[idx]);
}
sol.Maxflow(SS, ST, INF);
sol.AddEdge(T, S, INF);
sol.Maxflow(SS, ST, INF);
return sol.edges[sol.edges.size() - 2].flow;
} int main()
{
int T;
scanf("%d", &T);
while(T --)
{
scanf("%d%d", &n, &m);
n --;
scanf("%d%d", &p[0].x, &p[0].y);
p[0].b = p[0].e = 0;
for(int i = 1; i <= n; i ++)
p[i].inpt();
int ans = 0;
for(int i = 0; i < m; i ++)
ans += solve(i);
printf("%d\n", ans);
}
}