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：

弄清楚題意比較重要：

出租車公司有n個預定, 每一個預定有時間和地點, 地點分佈在二維整數座標系上, 地點之間的行駛時間爲兩點間的曼哈頓距離（|x1 - x2| + |y1 - y2|）。一輛車能夠在運完一個乘客後運另外一個乘客, 條件是此車要在預定開始前一分鐘以前到達出發地, 問最少須要幾輛車搞定全部預定。(摘自http://blog.sina.com.cn/s/blog_6635898a0100m54w.html)

 

我就是沒弄清題意WA了好幾回。弄清提議後開始構圖，求最小邊覆蓋，還有就是這是有向圖有因此構單邊。

 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 }