Taxi Cab Scheme

Description

Running a taxi station is not all that simple. Apart from the obvious demand for a centralised coordination of the cabs in order to pick up the customers calling to get a cab as soon as possible,there is also a need to schedule all the taxi rides which have been booked in advance.Given a list of all booked taxi rides for the next day, you want to minimise the number of cabs needed to carry out all of the rides.
For the sake of simplicity, we model a city as a rectangular grid. An address in the city is denoted by two integers: the street and avenue number. The time needed to get from the address a, b to c, d by taxi is |a - c| + |b - d| minutes. A cab may carry out a booked ride if it is its first ride of the day, or if it can get to the source address of the new ride from its latest,at least one minute before the new ride's scheduled departure. Note that some rides may end after midnight.

Input

On the first line of the input is a single positive integer N, telling the number of test scenarios to follow. Each scenario begins with a line containing an integer M, 0 < M < 500, being the number of booked taxi rides. The following M lines contain the rides. Each ride is described by a departure time on the format hh:mm (ranging from 00:00 to 23:59), two integers a b that are the coordinates of the source address and two integers c d that are the coordinates of the destination address. All coordinates are at least 0 and strictly smaller than 200. The booked rides in each scenario are sorted in order of increasing departure time.

Output

For each scenario, output one line containing the minimum number of cabs required to carry out all the booked taxi rides.

Sample Input

2

2

08:00 10 11 9 16

08:07 9 16 10 11

2

08:00 10 11 9 16

08:06 9 16 10 11

Sample Output

1

2

Source

Northwestern Europe 2004

       根據這道題目的意思，我們可以建一張圖，對于兩個booked taxi ride，ri和rj如果一輛車能夠先完成ri的任務再有時間趕去完成rj的任務，那么就建立一條ri指向rj的邊。

       按照題目的要求，要選擇最少的taxi來完成這些任務。顯然在上面這個例子中，需要安排2輛taxi。結合這個圖，可以把題目的要求轉化為找出最少的路徑條數，使得這些路徑覆蓋途中所有的邊，例如可以選擇2條路徑1->3->4和1->2就可以覆蓋所有的邊。也可以選擇1->3->4和2（因為2作為初始站，不需要由1轉移過來）。對于一條連續的路徑vi1->vi2->…vik由于這條路徑上的任務實際上是由一輛taxi來完成的，可以吧這條路徑退化成兩個點vi1->vik。有了這兩步建圖的步驟以后，問題的求解就可以變為找出頂點集的一個最小子集，使這個頂點子集覆蓋所有的邊（每條邊都至少和一個頂點集的頂點相連）。這個問題就是圖的最小點覆蓋。再看這張圖，還有一些性質能夠讓我們更好地求出最小點覆蓋。這個圖是一個有向無環圖，沒有自環，就可以拆點，把原先建的圖變成一張二分圖。

可以再圖中看出，上面舉出的一條路徑1->3->4對應了這個二分圖中的路徑1->3’->3->4’，在這個二分圖中就需要求一個最大獨立子集（這里的4點就是一條路徑的終點，沒一條路徑即對應有一個終點！）。二分圖的最大獨立數是總點數與最大匹配數的差值。接下來建圖，拆點，求二分圖最大匹配就能解決這道題目了。

 1#include <stdio.h>
 2#include <string.h>
 3#include <algorithm>
 4using namespace std;
 5
 6struct Point
 7{
 8    int x,y;
 9};
10struct P
11{
12    int s,e;
13    struct Point s1,s2;
14}p[505];
15int n,g[505][505],match[505];
16bool chk[505];
17
18/**//*int abs(int a)
19{
20    return a < 0 ? -a : a;
21}*/
22bool operator <(const P a, const P b)
23{
24    return a.s < b.s;
25}
26int dfs(int v)
27{
28    int i,t;
29    for(i=0; i<n; i++)
30    {
31        if(!chk[i] && g[v][i])
32        {
33            t=match[i];
34            chk[i]=true;
35            match[i]=v;
36            if(t == -1 || dfs(t)) return 1;
37            match[i]=t;
38        }
39    }
40    return 0;
41}
42int Twomatch()
43{
44    int i,ans=0;
45    memset(match,-1,sizeof(match));
46    for(i=0; i<n; i++)
47    {
48        memset(chk,false,sizeof(chk));
49        if(dfs(i)) ans++;
50    }
51    return ans;
52}
53
54int main()
55{
56    int i,cases,a,b,j,cnt;
57    char temp[10];
58    //freopen("in.txt","r",stdin);
59    scanf("%d",&cases);
60    while(cases--)
61    {
62        scanf("%d",&n);
63
64        for(i=0; i<n; i++)
65        {
66            scanf("%s",temp);
67            scanf("%d%d%d%d",&p[i].s1.x,&p[i].s1.y,&p[i].s2.x,&p[i].s2.y);
68            sscanf(temp,"%d%*[:]%d",&a,&b);
69            p[i].s=a*60+b;
70            p[i].e=p[i].s+abs(p[i].s2.x-p[i].s1.x)+abs(p[i].s2.y-p[i].s1.y);
71        }
72        //sort(p,p+n);
73        cnt=0;
74        memset(g,0,sizeof(g));
75        for(i=0; i<n; i++)
76        {
77            for(j=0; j<n; j++)
78            {
79                if(p[i].e+abs(p[i].s2.x-p[j].s1.x)+abs(p[i].s2.y-p[j].s1.y) < p[j].s && i != j)
80                    g[i][j]=1;
81            }
82        }
83        printf("%d\n",n-Twomatch());
84    }
85    return 0;
86}
87