如果一個計程車在接完一個客人之後還可以接另一位客人，那這兩點之間連一條邊。

現在突然有個疑問，為什麼不用求Floyd？好吧，我明白了，建圖過程中就把所有邊都連上了，相當於求了一遍Floyd~

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>

using namespace std;

const int MAX_N = 500 + 10;

struct Node
{
    int t1, t2, x1, y1, x2, y2;
};

struct Edge
{
    int v, next;
};

Edge e[MAX_N * (MAX_N - 1)];
Node node[MAX_N];
bool vis[MAX_N];
int head[MAX_N], link[MAX_N];
int n, cnt;

void addEdge(int u, int v)
{
    e[cnt].v = v;
    e[cnt].next = head[u];
    head[u] = cnt++;
}

bool DFS(int u)
{
    int v;
    for(int i = head[u]; i != -1; i = e[i].next)
    {
        v = e[i].v;
        if(!vis[v])
        {
            vis[v] = 1;
            if(link[v] == -1 || DFS(link[v]))
            {
                link[v] = u;
                return true;
            }
        }
    }
    return false;
}

int MaxMatch()
{
    int ans = 0;
    memset(link, -1, sizeof(link));
    for(int i = 0; i < n; i++)
    {
        memset(vis, 0, sizeof(vis));
        if(DFS(i))
            ans++;
    }
    return ans;
}

int dis(int x1, int y1, int x2, int y2)
{
    return abs(x2 - x1) + abs(y2 - y1);
}

int main()
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        cnt = 0;
        memset(head, -1, sizeof(head));
        scanf("%d", &n);
        int h, m;
        for(int i = 0; i < n; i++)
        {
            scanf("%d:%d%d%d%d%d", &h, &m, &node[i].x1, &node[i].y1, &node[i].x2, &node[i].y2);
            node[i].t1 = h * 60 + m;
            node[i].t2 = node[i].t1 + dis(node[i].x1, node[i].y1, node[i].x2, node[i].y2);
        }
        for(int i = 0; i < n; i++)
            for(int j = i + 1; j < n; j++)
                if(node[i].t2 + dis(node[i].x2, node[i].y2, node[j].x1, node[j].y1) + 1 <= node[j].t1)
                    addEdge(i, j);
        printf("%d\n", n - MaxMatch());
    }
    return 0;
}