POJ 2060 最小路径覆盖

题目：
Taxi Cab Scheme
Time Limit: 1000MS Memory Limit: 30000K
Total Submissions: 4432 Accepted: 1868
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：


有一些出租车，分别要从一个地方去往另一个地方载客，从一个地方驶向另一个地方的时间为这两个地方的行坐标之差的绝对值+列坐标之差的绝对值，如果一个出租车从一个地方达到另一个地方然后驶向另一个起点时间足够的话，可以取代另一个出租车，这样就可以少用一个出租车，题目求的是最多可以少用几辆出租车。
给定测试数据组数，每组给定出租车数量，然后是没个出租车的出发时间，起点坐标，终点坐标。


源代码：


#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;


struct texi
{
int s_time;
int e_time;
int br;
int bc;
int er;
int ec;
};


int n;
texi tx[600];
int a[600][600];
int a1[600];
int a2[600];
bool fl[600];


int inline Abs(int a)
{
return a>0?a:-a;
}


bool search(int v)
{
for(int i=1;i<=n;i++)
{
if(a[v][i]&&!fl[i])
{
fl[i]=1;
if(a2[i]==-1||search(a2[i]))
{
a2[i]=v;
a1[v]=i;
return 1;
}
}
}
return 0;
}


int maxmatch()
{
memset(a1,-1,sizeof(a1));
memset(a2,-1,sizeof(a2));
int ret=0;
for(int i=1;i<=n;i++)
{
if(a1[i]==-1)
{
memset(fl,0,sizeof(fl));
if(search(i))
{
ret++;
}
}
}
return ret;
}


int main()
{
int t;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
for(int i=1;i<=n;i++)
{
int h,m;
scanf("%d:%d%d%d%d%d",&h,&m,&tx[i].br,&tx[i].bc,&tx[i].er,&tx[i].ec);
int tim=Abs(tx[i].br-tx[i].er)+Abs(tx[i].bc-tx[i].ec);
tx[i].s_time=h*60+m;
tx[i].e_time=tx[i].s_time+tim;
}
memset(a,0,sizeof(a));
for(int i=1;i<=n;i++)
{
for(int j=1;j<=n;j++)
{
if(j==i)continue;
if(tx[j].s_time>tx[i].e_time+Abs(tx[i].er-tx[j].br)+Abs(tx[i].ec-tx[j].bc))
{
a[i][j]=1;
}
}
}
/*
for(int i=1;i<=n;i++)
{
for(int j=1;j<=n;j++)
{
printf("%d ",a[i][j]);
}printf("\n");
}
*/
int res=maxmatch();
printf("%d\n",n-res);
}
return 0;
}