POJ 2060 Taxi Cab Scheme (最小路径覆盖)

Taxi Cab Scheme

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 5148		Accepted: 2150

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

题意：给你N个出租车的预定单表，有初始时间，起点和终点。问最少用多少辆出租车可以满足这N个预订单。

思路：根据题意就可以建出图，对于每个任务ride_i来说，完成这个任务的出租车如果完成之后还能有时间去完成其他的ride_j，那就从ride_i到ride_j连接一条有向边。然后会发现求最少的出租车数其实就是求最小路径覆盖。刚开始我一直认为ride_i的终点要和ride_j的起点一样才行，结果一直WA,看来我太天真了.....其实只要出租车i完成自己的任务后，在开到j的起点，只要在它的发车时间前到达就行，既可以连接一条边.当然对于时间的计算我有点麻烦了，其实可以把所有的时间都化成time=hours*60+minutes的形式

另外，最小路径覆盖=总匹配数量N-最大匹配数。

#include <cstdio>
#include <cstring>
#include <vector>
#include <iostream>
using namespace std;
const int maxn=505;
vector<int>edge[maxn];
bool vis[maxn];
int match[maxn],p;
struct node{
int ho,mi;
int tho,tmi;
int s[2],t[2];
}ride[maxn];
inline int f(int a)
{
    return a<0?(-a):a;
}
bool ok(node a,node b)
{
    if(a.tho*60+a.tmi+f(a.t[0]-b.s[0])+f(a.t[1]-b.s[1])-b.ho*60-b.mi<=0)
    return true;
    return false;
}
bool find(int now)
{
    int i,v;
    for(i=0; i<edge[now].size(); ++i)
        if(!vis[v=edge[now][i]])
        {
            vis[v]=true;
            if(match[v]==-1||find(match[v]))
            {
                match[v]=now;
                return true;
            }
        }
    return false;
}
int main()
{
    int i,j,m,tmp;
    int ca;
    scanf("%d",&ca);
    while(ca--)
    {
        scanf("%d",&m);
        memset(match,-1,sizeof(match));
        for(i=1; i<=m; i++)
            {
                edge[i].clear();
                scanf("%d:%d",&ride[i].ho,&ride[i].mi);
                scanf("%d%d%d%d",&ride[i].s[0],&ride[i].s[1],&ride[i].t[0],&ride[i].t[1]);
                tmp=f(ride[i].s[0]-ride[i].t[0])+f(ride[i].s[1]-ride[i].t[1]);
                ride[i].tmi=(ride[i].mi+tmp+1)%60;
                ride[i].tho=(ride[i].mi+tmp+1)/60+ride[i].ho;
            }
        for(i=1;i<=m;i++)
        for(j=1;j<=m;j++)
        if(ok(ride[i],ride[j]))
            edge[i].push_back(j);
        int res=0;
        for(i=1; i<=m; i++)
            {
                memset(vis,0,sizeof(vis));
                if(find(i)) res++;
            }
        printf("%d\n",m-res);
    }
    return 0;
}