HDU 1350 & HDU 1960 & POJ 2060 Taxi Cab Scheme【二分图之最小路径覆盖，经典】

Taxi Cab Scheme

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 310    Accepted Submission(s): 195

Problem 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

NWERC2004

原题链接：http://acm.hdu.edu.cn/showproblem.php?pid=1960

题意：样例1中: 8:00的时候要把第一位乘客从（10,11）送到（9,16），花费时间为6min（即曼哈顿距离）。

这辆车可以在（9,16）等到08:07 的时候送第二名乘客到达目的地。

而样例2中，第二名乘客08:06就要出发，而第一辆车08:06才到，所以要重新派一辆车。因为题目中说"The booked rides in each scenario are sorted in order of increasing departure time."

这道题很难想到是二分匹配的最小路径覆盖。

想到的话建图也是关键。

再写一遍公式：

最小路径覆盖=顶点数 - 最大匹配数。

具体细节看代码吧。

AC代码：

#include <cstdio>
#include <cstring>
#include <cmath>
#include <iostream>
using namespace std;
int n;
const int maxn=500+5;
bool a[maxn][maxn];
bool vis[maxn];
int link[maxn];
struct Node
{
    int sx,sy;//起点
    int ex,ey;//终点
    int s,e;//到达时间，离开时间
}node[maxn];
bool Find(int x)
{
    for(int i=0;i<n;i++)
    {
        if(!vis[i]&&a[x][i])
        {
            vis[i]=true;
            if(!link[i]||Find(link[i]))
            {
                link[i]=x;
                return true;
            }
        }
    }
    return false;
}
int Hungery()
{
    int ans=0;
    memset(link,0,sizeof(link));
    for(int i=0;i<n;i++)
    {
        memset(vis,false,sizeof(vis));
        if(Find(i))
            ans++;
    }
    return ans;
}
int main()
{
    int T;
    //freopen("data/1350.txt","r",stdin);
    //freopen("data/1350_me.txt","w",stdout);
    cin>>T;
    while(T--)
    {
        cin>>n;
        int h,m,sx,sy,ex,ey;
        for(int i=0;i<n;i++)
        {
            scanf("%d:%d",&h,&m);
            node[i].s=h*60+m;
            cin>>sx>>sy>>ex>>ey;
            node[i].sx=sx,node[i].sy=sy;
            node[i].ex=ex,node[i].ey=ey;
            node[i].e=node[i].s+abs(sx-ex)+abs(sy-ey);
        }
        memset(a,false,sizeof(a));
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<n;j++)
            {
                int t=abs(node[i].ex-node[j].sx)+abs(node[i].ey-node[j].sy);
                //第i个点的结束时间+到达第j个点的时间<第j个点的开始时间
                if(i!=j&&node[i].e+t<node[j].s)
                {
                    a[i][j]=true;
                }
            }
        }
        cout<<n-Hungery()<<endl;
    }
    return 0;
}

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