poj 2060 Taxi Cab Scheme (二分匹配)




Taxi Cab Scheme










Time Limit: 1000MS   Memory Limit: 30000K
Total Submissions: 5710   Accepted: 2393

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

二分匹配:

弄清楚题意比较重要:

出租车公司有n个预约, 每个预约有时间和地点, 地点分布在二维整数坐标系上, 地点之间的行驶时间为两点间的曼哈顿距离(|x1 - x2| +
|y1 - y2|)。一辆车可以在运完一个乘客后运另一个乘客, 条件是此车要在预约开始前一分钟之前到达出发地, 问最少需要几辆车搞定所有预约。(摘自http://blog.sina.com.cn/s/blog_6635898a0100m54w.html)

我就是没弄清题意WA了好几次。弄清提议后开始构图,求最小边覆盖,还有就是这是有向图有所以构单边。


 1 //692K    79MS    C++    1333B    2014-06-05 11:26:44
2 #include<iostream>
3 #include<vector>
4 #define N 505
5 using namespace std;
6 struct node{
7 int a,b,c,d;
8 int time;
9 }p[N];
10 vector<int>V[N];
11 int match[N];
12 int vis[N];
13 int n;
14 int dfs(int u)
15 {
16 for(int i=0;i<V[u].size();i++){
17 int v=V[u][i];
18 if(!vis[v]){
19 vis[v]=1;
20 if(match[v]==-1 || dfs(match[v])){
21 match[v]=u;
22 return 1;
23 }
24 }
25 }
26 return 0;
27 }
28 int hungary()
29 {
30 int ret=0;
31 memset(match,-1,sizeof(match));
32 for(int i=0;i<n;i++){
33 memset(vis,0,sizeof(vis));
34 ret+=dfs(i);
35 }
36 return ret;
37 }
38 int main(void)
39 {
40 int t;
41 int time[N];
42 int dis[N];
43 int h,m;
44 scanf("%d",&t);
45 while(t--)
46 {
47 scanf("%d",&n);
48 for(int i=0;i<=n;i++) V[i].clear();
49 for(int i=0;i<n;i++){
50 scanf("%d:%d %d%d%d%d",&h,&m,&p[i].a,&p[i].b,&p[i].c,&p[i].d);
51 p[i].time=abs(p[i].a-p[i].c)+abs(p[i].b-p[i].d);
52 time[i]=h*60+m;
53 for(int j=0;j<i;j++)
54 if(time[i]-time[j]>p[j].time+abs(p[i].a-p[j].c)+abs(p[i].b-p[j].d)){
55 //V[i].push_back(j);
56 V[j].push_back(i);
57 }
58 }
59 printf("%d\n",n-hungary());
60 }
61 return 0;
62 }

poj 2060 Taxi Cab Scheme (二分匹配),布布扣,bubuko.com

时间: 2024-10-12 13:20:11

poj 2060 Taxi Cab Scheme (二分匹配)的相关文章

POJ 2060 Taxi Cab Scheme【最小路径覆盖】

T - Taxi Cab Scheme Time Limit:1000MS     Memory Limit:30000KB     64bit IO Format:%I64d & %I64u Submit Status Practice POJ 2060 Appoint description:  System Crawler  (2014-08-22) Description Running a taxi station is not all that simple. Apart from

poj 2060 Taxi Cab Scheme 最小路径覆盖

//二分匹配的最小路径覆盖 //对于第i次ride,如果在第i次ride结束后还能在第j次ride出发前赶到第j次的出发点 //那么i到j就有一条边 //根据最小路径覆盖 = N - 最大匹配即可得到答案 #include<iostream> #include<cstdio> #include<cstring> #include<cstdlib> using namespace std; const int maxn  = 510; int line[max

poj 2060 Taxi Cab Scheme(DAG图的最小路径覆盖)

题意: 出租车公司有M个订单. 订单格式:     hh:mm  a  b  c  d 含义:在hh:mm这个时刻客人将从(a,b)这个位置出发,他(她)要去(c,d)这个位置. 规定1:从(a,b)到(c,d)所花的时间为:abs(a-c)+abs(b-d). 规定2:一辆出租车如果要接单,必须在客人出发前1分钟(包括)以前接单. 问,最少派出多少辆出租车,可以完成所有任务. 思路: 把每一笔单看成一个点,如果完成第i个单后可以接到第j个单,则 i->j连上线. 则题为:求这个DAG图的最小路

POJ - 2060 Taxi Cab Scheme 二分图 最小路径覆盖

题目大意:有n项任务,给出每项任务的出发时间,出发地点和目的地. 当一个任务完成之后,如果 当前时间 + 到达另一个任务的出发地的时间 <= 另一个任务的出发时间 - 1 的话,那么就可以让这个人接下去完成这个任务了 问至少需要派多少人才可以完成任务 解题思路:这题和 poj-3216 Repairing Company很像 两个点集都是任务,如果一个任务完成了可以接下来完成另一个任务的话,那么这两个任务之间就存在关系了,这样二分图就建成了 因为要把所有的任务都完成,也就是说要覆盖所有点,这就变

hdu 1350 Taxi Cab Scheme(二分匹配)

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1350 Taxi Cab Scheme Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 943    Accepted Submission(s): 456 Problem Description Running a taxi stati

UVA 1201 - Taxi Cab Scheme(二分图匹配+最小路径覆盖)

UVA 1201 - Taxi Cab Scheme 题目链接 题意:给定一些乘客,每个乘客需要一个出租车,有一个起始时刻,起点,终点,行走路程为曼哈顿距离,每辆出租车必须在乘客一分钟之前到达,问最少需要几辆出租车 思路:如果一辆车载完一个乘客a,能去载乘客b,就连一条有向边,这样做完整个图形成一个DAG,然后要求的最少数量就是最小路径覆盖,利用二分图最大匹配去做,把每个点拆成两点,如果有边就连边,做一次最大匹配,n - 最大匹配数就是答案 代码: #include <cstdio> #inc

二分图最小路径覆盖--poj2060 Taxi Cab Scheme

Taxi Cab Scheme 时间限制: 1 Sec  内存限制: 64 MB 题目描述 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

POJ 3014:Asteroids(二分匹配,匈牙利算法)

Asteroids Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 14399   Accepted: 7836 Description Bessie wants to navigate her spaceship through a dangerous asteroid field in the shape of an N x N grid (1 <= N <= 500). The grid contains K as

poj 1274 The Perfect Stall (二分匹配)

The Perfect Stall Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 17768   Accepted: 8104 Description Farmer John completed his new barn just last week, complete with all the latest milking technology. Unfortunately, due to engineering pr